Proof For $\Sigma $ -types, let us consider a dependent tuple (see the Appendix for definition) $(\Gamma , (A, A[-]), (B, B[-]))$ of $(\rho ^R, \chi ^R)$ . The $\Sigma $ -type associated with this tuple has the following form: $(\Sigma _A B, \Sigma _A B [-])$ where $\Sigma _A B$ is the $\Sigma $ -type given by the pseudo-stable choice of $(\rho , \chi )$ applied to $(\Gamma , A, B)$ . The component at $\sigma \colon \Delta \rightarrow \Gamma $ of the local cleavage $\Sigma _A B [-]$ is given as follows. First, we use the local cleavages $A[-]$ and $B[-]$ to construct a Cartesian arrow of dependent tuples $(\sigma , f^*, g^*) \colon (\Delta , A[\sigma ], B[\sigma ]) \rightarrow (\Gamma , A, B)$ where the arrows $f^*$ and $g^*$ are given by the local cleavage (see Definition 2.3). Then, we use the action on morphisms of the pseudo-stable choice of $\Sigma $ -types to define:

This local cleavage is normal because the pseudo-stable choice is functorial.

We must show that this choice is strictly stable. By definition, for $\sigma \colon \Delta \rightarrow \Gamma $ ,

$$ \begin{align*} (\Sigma_A B , \Sigma_A B [-]) [\sigma] &= ((\Sigma_{A} B)[\sigma], (\Sigma_{A} B)[\sigma][-]) \quad (\text{by def.~of the cleavage of } (\rho^R, \chi^R))\\ &= (\Sigma_{A[\sigma]} B[\sigma], (\Sigma_{A} B)[\sigma][-]) \quad (\text{by def.~of } \Sigma_A B [-] ) \rlap{.} \end{align*} $$

It only remains to show that the local cleavages $(\Sigma _A B)[\sigma ][-]$ and $(\Sigma _{A[\sigma ]} B[\sigma ]) [-]$ coincide, but this follows from the functoriality of the pseudo-stable choice of $\Sigma $ -types in $(\rho , \chi )$ .

The construction for the case of dependent products or $\Pi $ -types is completely analogous and hence omitted.

For $\textsf {Id}$ -types, note that the terms of a type $(A, A[-])$ in $(\mathbb {C}^R, \rho ^R, \chi ^R)$ are the same as the terms of A in $(\mathbb {C}, \rho , \chi )$ . Let $(A, A[-])$ be an object in the fibre of $\rho ^R$ over $\Gamma $ , and consider sections $a, b \colon \Gamma \to \Gamma .A$ . We need an object in $\mathbb {E}^R$ over $\Gamma .A.A$ , which we denote as

$$ \begin{align*} \big( {\textsf{Id}}_{A}(a, b), {\textsf{Id}}_A(a, b)[-] \big) \end{align*} $$

for brevity. The object ${\textsf {Id}}_A(a, b)$ is obtained by applying the pseudo-stable choice of $\textsf {Id}$ -types in $(\rho , \chi )$ to A and the sections $a, b \colon \Gamma \to \Gamma .A$ .

To define the local normal cleavage ${\textsf {Id}}_{A}(a, b)[-]$ , consider $\sigma \colon \Delta \rightarrow \Gamma $ in $\mathbb {C}$ . The local normal cleavage $A[-]$ gives a Cartesian arrow $\sigma ^* \colon A[\sigma ] \rightarrow A$ over $\sigma $ . Using the stable functorial choice of $\textsf {Id}$ -types of $(\rho , \chi )$ , we get the Cartesian arrow

$$ \begin{align*} {\textsf{Id}}_{\sigma^*}(a, b) \colon {\textsf{Id}}_{A[\sigma]}(a[\sigma], b[\sigma]) \rightarrow {\textsf{Id}}_A(a, b) \end{align*} $$

over $\sigma $ . We then let ${\textsf {Id}}_{A}(a, b)[\sigma ] := {\textsf {Id}}_{A[\sigma ]}(a[\sigma ], b[\sigma ])$ , so that the Cartesian arrow to ${\textsf {Id}}_A(a, b)$ can be taken to be ${\textsf {Id}}_{\sigma ^*}(a, b)$ . By pseudo-stability, if $\sigma = 1_A$ then

$$ \begin{align*} {\textsf{Id}}_{A[\sigma]}(a[\sigma], b[\sigma]) = {\textsf{Id}}_{A}(a, b) \, \quad {\textsf{Id}}_{\sigma^*}(a, b) = 1_{{\textsf{Id}}_{A}(a, b)} \rlap{,} \end{align*} $$

as required. It remains to check that this choice is strictly stable. For $\sigma \colon \Delta \rightarrow \Gamma $ and sections $a, b \colon \Gamma \to \Gamma .A$ , we must verify that:

$$ \begin{align*} ({\textsf{Id}}_A(a, b)[\sigma], {\textsf{Id}}_A(a, b)[\sigma][-]) = ({\textsf{Id}}_{A[\sigma]}(a[\sigma], b[\sigma]), {\textsf{Id}}_{A[\sigma]}(a[\sigma], b[\sigma]) [-]). \end{align*} $$

By definition, ${\textsf {Id}}_{A}(a, b)[\sigma ]$ is given by the local normal cleavage ${\textsf {Id}}_A[-]$ applied to the arrow $\sigma ^*$ , which is ${\textsf {Id}}_{A[\sigma ]}(a[\sigma ], b[\sigma ])$ . By the functoriality of the pseudo-stable choice of $\textsf {Id}$ -types, the local cleavages ${\textsf {Id}}_A(a, b)[\sigma ][-]$ and ${\textsf {Id}}_{A[\sigma ]}(a[\sigma ], b[\sigma ]) [-]$ coincide.

Proof Suppose we have a pseudo-stable choice of $\textsf {Id}$ -types. To construct a pseudo-stable choice of variable-based $\textsf {Id}$ -types, consider an object A over $\Gamma $ , and we define ${\textsf {Id}}_A := {\textsf {Id}}_A(x, y)$ over $\Gamma .A.A$ using the canonical variables $x, y \colon \Gamma .A.A \to \Gamma .A.A.A$ . The operations r and j are given just as in Definition 2.5.

For the converse, assume a pseudo-stable choice of variable-based $\textsf {Id}$ -types. Consider an object A over $\Gamma $ and sections $a, b \colon \Gamma \to \Gamma .A$ . We have ${\textsf {Id}}_A$ over $\Gamma .A.A$ , and thus, we can define ${\textsf {Id}}_A(a, b)$ to be the reindexing of ${\textsf {Id}}_A$ along $(a, b)\colon \Gamma \to \Gamma .A.A$ , as in

(2.1)

We can extend this definition to provide the rest of the data in part $(1)$ of Definition 2.5. Consider $f \colon B \to A$ Cartesian over $\sigma \colon \Delta \to \Gamma $ , then we have

where ${\textsf {Id}}_f(a, b) \colon {\textsf {Id}}_{B}(a[\sigma ], b[\sigma ]) \rightarrow {\textsf {Id}}_A(a, b)$ is given uniquely by the universal property of the square in (2.1). If f (and $\sigma $ ) are identities, then by pseudo-stability of $\textsf {Id}$ -types, ${\textsf {Id}}_f \colon {\textsf {Id}}_A \to {\textsf {Id}}_A$ is also the identity, and thus ${\textsf {Id}}_f(a, b)$ must be the identity by the uniqueness property that characterises it. Similarly, the pseudo-stability of the variable-based $\textsf {Id}$ -types and the uniqueness property of ${\textsf {Id}}_f(a, b)$ can be used to show that this operation preserves composition.

For the operations r and j, we first consider the diagram

where the square on the right is obtained by the functoriality of the pseudo-stable choices of variable-based $\textsf {Id}$ -types to $\chi _{A, A} \colon \Gamma .A.A \to \Gamma $ . The square on the left is obtained by definition of ${\textsf {Id}}_A(x, y)$ applied to the object A weakened to $\Gamma .A.A$ and to the variables $x, y \colon \Gamma .A.A \to \Gamma .A.A.A$ . Both top horizontal arrows are Cartesian and the composition of the bottom two arrows equals the identity. Thus ${\textsf {Id}}_A(x, y) \cong {\textsf {Id}}_A$ as objects over $\Gamma .A.A$ . We can then transport the operations r and j along this isomorphism.

Proof Let $(f, s) \colon X \to \Gamma $ and $(g, t) \colon Y \to X$ be in $\textbf{R}\text{-}\textbf{Map}$ . The pullback functor along $f \colon X \to \Gamma $ has a left adjoint $\Sigma _f \colon \mathbb {C}/X \rightarrow \mathbb {C}/\Gamma $ , which is given by composition. By Remark 3.2, this functor lifts to slices of $\textbf{R}\text{-}\textbf{Map}$ , as follows:

For the formation rule, we define $\Sigma _f g \colon = \Sigma _f (g) = f\circ g\colon Y \to \Gamma $ . For the introduction rule, we define $\textrm{pair}_{f,g} \colon Y \rightarrow Y$ over f,

by letting $\textrm{pair}_{f,g} \colon = 1_{Y}$ . Finally, for the elimination rule, let C be over $\Sigma _f g$ and let t be a section of C over $\textrm{pair}_{f, g}$ , and we define $\textrm{sp}_{f,g}(C,t) \colon = t$ . The computation rule holds trivially, thus giving rise to a choice of $\Sigma $ -types. Stable functoriality and coherence of elimination terms also follow easily; the crucial observation is that vertical composition of R-maps plays nicely with the horizontal categorical structure (see [Reference Bourke and Garner5, Section 2.8]).

Proof Consider $(f, s) \colon X \rightarrow \Gamma $ in $\textbf{R}\text{-}\textbf{Map}$ . By the exponentiability property, we have a pushforward functor $\Pi _f \colon \underline {\text{R}}/X \to \mathbb {C}/\Gamma $ (here $\underline {\text{R}}/X$ denotes the slice category whose objects are arrows $g \colon Y \to X$ that can be equipped with an R-map structure) and, by [Reference Gambino and Sattler14, Proposition 6.5 and Proposition 6.7] this lifts to a functor

Using the functors

of Proposition 3.3 we can find a lift of $\Pi _f$ as follows:

Consider an R-map $(g ,t) \colon Y \to X$ . For the formation rule, we apply $\Pi _f$ to obtain an R-map $\Pi _f g \colon Y \rightarrow \Gamma $ . For the introduction rule, we define the operation $\lambda $ ; consider a section t of g, this is an arrow $t \colon 1_{X} \rightarrow g$ in $\mathbb {C}/X$ and since $f^*(1_{\Gamma }) \cong 1_{X}$ this is the same thing as an arrow $t \colon f^*(1_{\Gamma }) \rightarrow g$ . Taking the transpose yields a map $\lambda (t) \colon 1_{\Gamma } \rightarrow \Pi _f g$ , as required. For the elimination rule we need to provide an arrow $\textrm{app}_{f,g} \colon f^*(\Pi _f g) \rightarrow g$ . We can take $\textrm{app}_{f,g}$ to be the counit of the adjunction, $\textrm{app}_{f,g}\colon = \epsilon _{g} \colon f^*(\Pi _f g) \rightarrow g$ . The computation rule follows easily from the bijection $\lambda \colon \mathbb {C} /X [1_{X}, g] \to \mathbb {C} /\Gamma [1_{\Gamma }, \Pi _f g]$ .

It only remains to show the assignment $(\Gamma , f, g) \mapsto (\Pi _f g, \lambda ,\ \textrm{app})$ is pseudo-stable. This is a diagram-chasing argument, which relies on the fact that, for a Cartesian square of the form

the Beck-Chevalley isomorphism $BC \colon \Delta _\sigma \Pi _f \rightarrow \Pi _{f'} \Delta _\tau $ (where we write $\Delta _\sigma $ and $\Delta _\tau $ for the pullback functors along $\sigma $ and $\tau $ , respectively) lifts to an isomorphism of R-maps by [Reference Gambino and Sattler14, Proposition 6.7]. We leave the details to the readers.

Proof We first construct a pseudo-stable choice $({\textsf {Id}}_{v.b.}, r, j)$ of variable-based $\textsf {Id}$ -types (see Definition 2.7) and then apply Proposition 2.9 in order to obtain a pseudo-stable choice of $\textsf {Id}$ -types (see Definition 2.5).

The choices for $\textsf {Id}$ and r are canonically given by the stable functorial choice of path objects. These satisfy the coherence properties of Definition 2.8. Since the maps $r_f$ are equipped with an L-map structure, we have lifts against R-maps. Using this, we obtain a choice of canonical elimination terms (i.e., j-terms).

We are left to verify that this choice is coherent. For this, it is sufficient to show that given a Cartesian morphism of R-maps $(h , k) \colon f' \rightarrow f$ , a R-map $q \colon C \rightarrow PX$ , if the diagram on the left of (4.1) commutes, then so does the one on the right.

(4.1)

where $q^* \colon C^* \rightarrow PX'$ is defined as the pullback of q along $P(h,k)$ . The arrows denoted by j are the canonical choices of lifts. The arrow $d^*$ is the pullback of d along $P(h, k)$ , i.e., it is defined to be the unique arrow $d^* \colon X' \rightarrow C^*$ such that:

(4.2) $$ \begin{align} q^* \circ d^* = r_{f'} \quad \text{and} \quad P(h, k)^* \circ d^* = d \circ h. \end{align} $$

We split the problem into two. First, consider the following diagram equipped with the corresponding canonical lifts:

Note that $j = P(h, k)^* \circ j(d^*)$ since the Cartesian square $q^* \rightarrow q$ is a morphism of R-maps. Now consider the following lifting problem:

Once more, $j' = j(d) \circ P(h, k)$ since the square $r_{f'} \rightarrow r_f$ is morphism of L-maps. Finally, (4.2) tells us that the outer squares of the two previous diagrams are equal, implying that they have the same lift $j = j'$ . Thus, $P(h , k)^* \circ j(d^*) = j(d) \circ P(h, k)$ as needed.

Proof Apply Proposition 4.3, Proposition 4.6, and Proposition 4.9.

Proof Combine Theorem 2.6 and Theorem 4.11.

Proof We start by examining the structures of the $C_t$ -maps and the F-maps. We know that an F-map structure on a map $f \colon X \rightarrow Y$ corresponds to a lift s as shown on the diagram on the left below:

A closer analysis will show that s equips $f \colon X \rightarrow Y$ with the structure of a normal isofibration. An L-map structure on $g \colon A \rightarrow B$ is given by a lift $\lambda $ as shown on the diagram on the right of the previous figure. The structure obtained from such a lift $\lambda $ can be decomposed as $\lambda (b) = (\lambda _1(b) , b , \lambda _2(b))$ where $\lambda _1 \colon B \rightarrow A$ corresponds to a retraction of g and $\lambda _2 \colon 1_B \rightarrow g \circ \lambda _1$ corresponds to a natural transformation constant on the image of f. This information corresponds to the structure of a strong deformation retraction.

We construct the corresponding structures of a comonad and a monad for $C_t$ and F respectively. We provide a brief description and leave the details to the reader. The comultiplication $\delta _f \colon \mathbf {\downarrow f} \rightarrow \mathbf {\downarrow C_t f}$ for $C_t$ is defined by letting

$$ \begin{align*} \delta_f \colon (a, b , p) \mapsto (a, (a, b, p), (1_a, p) \colon (a, b, p) \rightarrow (a, Fa, 1_{fa})) \rlap{.} \end{align*} $$

Similarly, the endofunctor F has a multiplication $\mu _f \colon \mathbf {\downarrow Ff} \rightarrow \mathbf {\downarrow f}$ given by

$$ \begin{align*} \mu_f \colon ((a, b , p), \tilde{b}, \tilde{p} \colon \tilde{b} \rightarrow b) \mapsto (a, \tilde{b}, p \circ \tilde{p}) \, .\end{align*} $$

Proof We show that pulling back a $C_t$ -map along an F-map is uniformly a $C_t$ -map. Consider $(g, \lambda ) \colon A \rightarrow Y$ a $C_t$ -map and $(f, s) \colon X \rightarrow Y$ an F-map. Let $g' \colon A \times _Y X \rightarrow X$ be the pullback of g along f. We define a $C_t$ -map structure $\lambda '$ on $g'$ which, by Proposition 5.1, corresponds to a strong deformation retraction $(g', \lambda _1', \lambda _2')$ . Using that f corresponds to a normal isofibration, we can find for each point $x \in X$ , a point $x' \in X$ and a lift $\lambda _2'(x)$ of $\lambda _2(fx)$ , as in

We define $\lambda _1'(x) = (\lambda _1(fx), x')$ ; the homotopy $\lambda _2' \colon 1 \rightarrow g' \circ \lambda _1'$ is defined using the top arrow in the previous diagram.

Proof For an F-map $(f,s) \colon X \rightarrow Y$ , we need to uniformly provide a $C_t$ -map structure to $r_f$ and an F-map structure to $\rho _f$ . Let us define $\lambda _1 := t_f \colon Pf \rightarrow X$ the canonical target map. We define the natural transformation $\lambda _2 \colon 1_{Pf} \rightarrow r_f \circ t_f$ by

$$ \begin{align*} \lambda_2(a, a', p) := (p, 1_{a'}) \colon (a, a', p) \rightarrow (a', a', 1_a') \rlap{.} \end{align*} $$

This corresponds to a strong deformation retraction structure on $r_f$ . An F-map structure on $\rho _f$ corresponds to a normal isofibration. Consider $(\alpha , \beta ) \colon (b, b') \rightarrow (a, a')$ in $X \times _Y X$ and an object $(a, a', p) \in Pf$ over $(a, a')$ . We find the lift $(\alpha , \beta ) \colon (b, b', q) \rightarrow (a, a', p)$ by setting $q := \beta \circ p \circ \alpha ^{-1} \colon b \rightarrow b'$ .

Proof Apply Proposition 5.1, Proposition 5.3, and Proposition 5.4.

Construction 6.2. Let us consider a suitable topos $(\mathbb {E}, I, \mathcal {M})$ . The category of arrows of trivial uniform fibrations, denoted by

$$ \begin{align*} \underline{\text{TrivUniFib}} \rightarrow \mathbb{E}^{\rightarrow} \rlap{,} \end{align*} $$

is defined as the right orthogonal category of arrow to $\mathcal {M}$ , that is, . Analogously, the category of arrows of uniform fibrations, denoted by

$$ \begin{align*} \underline{\text{UniFib}} \rightarrow \mathbb{E}^{\rightarrow} \rlap{,} \end{align*} $$

is defined as the right orthogonal category of arrow to $\mathcal {M}_{\hat {\times }}$ , i.e., .

Proof We start with $(1)$ . In order to define an inverse , consider an object , where $\theta $ is a lifting operation for squares $(a, b) \colon i \to f$ with $i \in \mathcal {I}$ . We need a way to canonically extend $\theta $ to all arrows in $\mathcal {M}$ . In order to do this, let us consider $j\colon A \to B$ in $\mathcal {M}$ . Since $\mathbb {A}$ is dense, we can express B canonically as a colimit $B \cong \textsf {colim}_{k \colon \mathscr {A} \to B} \mathscr {A}$ with $\mathscr {A} \in \mathbb {A}$ . Moreover, in a topos, pullbacks commute with colimits, and thus we obtain

$$ \begin{align*} j \cong \textsf{colim}_{k \colon \mathscr{A} \to B} ( \Delta_k j) \rlap{,} \end{align*} $$

where $\Delta _k j$ is the pullback of $j \colon A \to B$ along $k \colon \mathscr {A} \to B$ . Since $\mathcal {M}$ is closed under base change, $\Delta _k j \in \mathcal {M}$ and by definition, we get $ \Delta _k j \in \mathcal {I}$ . A filler for a square $(a, b) \colon j \to f$ is canonically obtained from the universal property of the colimit, applied to the collections of fillers given by $\theta $ relative to $\Delta _k j$ for each $k \colon \mathscr {A} \to B$ .

For $(2)$ , we proceed in a similar manner. Consider , we need to canonically extend $\theta $ to all arrows in $\mathcal {M}_{\hat {\times }}$ . For this, consider $\delta ^k \hat {\times } j \in \mathcal {M}_{\hat {\times }}$ . By definition, $j \in \mathcal {M}$ and thus $j \cong \textsf {colim}_{k \colon \mathscr {A} \to B} (k^*j)$ by the previous argument. Since $\delta ^k \hat {\times } -$ is cocontinuous, we obtain:

$$ \begin{align*} \delta^k \hat{\times} j \cong \delta^k \hat{\times} \textsf{colim}_{k \colon \mathscr{A} \to B} (\Delta_k j) \cong \textsf{colim}_{k \colon \mathscr{A} \to B} (\delta^k \hat{\times} ( \Delta_k j)) \rlap{,} \end{align*} $$

and by definition $\delta ^k \hat {\times } (\Delta _k ) \in \mathcal {I}_{\hat {\times }}$ . Once more, any square $(a, b) \colon \delta ^k \hat {\times } j \to f$ can be filled canonically by the universal property of the colimit applied to the collections of fillers given by $\theta $ relative to $\delta ^k \hat {\times } (\Delta _k j)$ .

Proof Apply Garner’s small object argument to $\mathcal {I}$ and $\mathcal {I}_{\hat {\times }}$ respectively. Since $\mathbb {E}$ is a Grothendieck topos, there exists a small dense subcategory $\mathbb {A}$ of $\mathbb {E}$ (for example, the full subcategory of compact objects for a large enough cardinal). By Lemma 6.3, the resulting awfs’s are algebraically-free on $\mathcal {M}$ and $\mathcal {M}_{\hat {\times }}$ respectively.

Proof Since the awfs of trivial uniform fibrations $(C, F_t)$ is suitable (see [Reference Gambino and Sattler14, Definition 7.1]), the functor $\delta ^k \hat {\times } - $ lifts to the category $\text{C}\text{-}\textbf{Map}$ and $\delta ^k \hat {\times } - $ also factors though the category $\underline{\text{SE}}$ of strong homotopy equivalences by [Reference Gambino and Sattler14, Lemma 8.4]. Combining these two facts, we obtain a lift

$$ \begin{align*} \delta^k \hat{\times} (-) \colon \text{C}\text{-}\textbf{Map} \to \text{C}\text{-}\textbf{Map} \times_{\mathbb{E}^{\rightarrow}} \underline{\text{SE}} \rlap{.} \end{align*} $$

By [Reference Gambino and Sattler14, Proposition 8.5], we have a functor $\text{C}\text{-}\textbf{Map} \times _{\mathbb {E}^{\rightarrow }} \underline {\text{SE}} \rightarrow \text{C}_{\text{t}}\text{-}\textbf{Map}$ over $\mathbb {E}^{\rightarrow }$ , and composing with the one above, we obtain a lift of $\delta ^k \hat {\times } - $

(6.3) $$ \begin{align} \delta^k \hat{\times} (-) \colon \text{C}\text{-}\textbf{Map} \to \text{C}_{\text{t}}\text{-}\textbf{Map} \rlap{.} \end{align} $$

By functorial orthogonality arguments with respect to the pushout-product and pullback-exponential constructions (see [Reference Gambino and Sattler14, Proposition 5.11] and [Reference Gambino and Sattler14, Remark 5.12]) together with the hypothesis (M5), the functor $i \hat {\times } - \colon \mathcal {M} \rightarrow \mathcal {M}$ lifts to the category $\text{C}\text{-}\textbf{Map}$ ,

(6.4) $$ \begin{align} i \hat{\times} (-) \colon \text{C}\text{-}\textbf{Map} \to \text{C}\text{-}\textbf{Map} \rlap{.} \end{align} $$

Applying the lifts in (6.3) and (6.4), together with the fact that $(C, F_t)$ is algebraically-free on the category of arrows $\mathcal {M} \rightarrow \mathbb {E}^{\rightarrow }$ , as witnessed by the functor $\eta \colon \mathcal {M} \rightarrow \text{C}\text{-}\textbf{Coalg}$ , we obtain the diagram

where $\tilde {\eta }$ is the composite of $\eta $ and the forgetful functor from C-algebras to C-maps.

By symmetry of the pushout-product functor, we obtain a natural isomorphism between $i \hat {\times } \delta ^k \hat {\times } -$ and $\delta ^k \hat {\times } i \hat {\times } -$ . We can transfer the algebraic structure along this natural isomorphism in order to obtain the following lift:

Taking the coproduct of these lifts for $k = 0, 1$ we obtain a lift of $i \hat {\times } -$ ,

(6.5) $$ \begin{align} {i \hat{\times} (-)} \colon \mathcal{M}_{\hat{\times}} \to \text{C}_{\text{t}}\text{-}\textbf{Map} \rlap{.} \end{align} $$

Using that

(cf. Proposition 3.3) and that $(C_t, F)$ is algebraically-free on $\mathcal {M}_{\hat {\times }}$ , we can apply [Reference Gambino and Sattler14, Proposition 5.9] to (6.5) and obtain

By the top pullback square in (6.2), the morphism $\rho _f \colon Pf \rightarrow B \times _A B$ is obtained in the following two steps:

$$ \begin{align*} f \mapsto \widehat{\hom}(i, f) \mapsto \langle \alpha_f, \lambda_f \rangle^*\widehat{\hom}(i, f) = \rho_f \rlap{,} \end{align*} $$

i.e., by first applying $\hat {\hom }(i, -)$ and then pulling back along $\langle \alpha _f, \lambda _f \rangle $ . Since we have lifts of $\hat {\hom }(i, -)$ and of the pullback functor to the category of F-algebras, we obtain a lift of $\rho $ ,

Finally, as we are working with an algebraically-free awfs, we have lifts back-and-forth between $\textbf{R}\text{-}\textbf{Alg}$ and $\textbf{R}\text{-}\textbf{Map}$ over $\mathbb {E}^{\rightarrow }$ (cf. Proposition 3.4), and thus we can transfer the lift of $\rho $ from the category of R-algebras to that of R-maps.

Proof We first show that the target map functor (that takes a map $f \colon B \rightarrow A$ to a map $t_{f} \colon Pf \rightarrow B$ ) lifts to a functor from $\text{F}\text{-}\textbf{Map}$ to $\text{F}_{\text{t}}\text{-}\textbf{Map}$ . Using that we have a lift $\delta ^1 \hat {\times } - \colon \text{C}\text{-}\textbf{Map} \rightarrow \text{C}_{\text{t}}\text{-}\textbf{Map}$ as shown in the proof of Lemma 6.6, we can transpose using [Reference Gambino and Sattler14, Proposition 5.9] to obtain a lift of $\hat{\hom}(\delta ^1, -)$ :

$$ \begin{align*} \widehat{\hom}(\delta^1, -) \colon \text{F}\text{-}\textbf{Alg} \to \text{F}_{\text{t}}\text{-}\textbf{Alg} \rlap{.} \end{align*} $$

Looking at (6.2), note that $t_f \colon Pf \rightarrow B$ is obtained by applying $\hat {\hom }(\delta ^1, -)$ to f and then pulling back along $\langle \beta _f, 1_B \rangle $ . Thus, the functor mapping $f \mapsto t_f$ lifts as

(6.6)

Since both awfs in question are algebraically-free, we can apply Proposition 3.4 to obtain the desired lift.

Let us return to the task of finding a lift of the functor $r \colon \mathbb {E}^{\rightarrow } \rightarrow \mathbb {E}^{\rightarrow }$ to a functor $r \colon \text{F}\text{-}\textbf{Map} \rightarrow \underline {\text{SDR}}$ . For this, we show that for each uniform fibration $(f, s) \colon B \rightarrow A$ the target map $t_f \colon Pf \rightarrow B$ is a strong homotopy retraction of $r_f \colon B \rightarrow Pf$ .

Looking again at (6.2) it is clear that $t_f \circ r_f = 1_B$ . Thus, we are left with the task of constructing an homotopy $H \colon r_f \circ t_f \sim 1_{P f}$ , for this consider the commutative diagram

where the top horizontal arrow is given by the universal property of $P f^{\partial I} \cong P f \times P f$ . This gives us an arrow into the pullback

$$ \begin{align*} \tilde{H} \colon P f \rightarrow B^I \times_{B^{\partial I}} P f^{\partial I}. \end{align*} $$

We already have a lift of the target map $t_{(-)} \colon \text{F}\text{-}\textbf{Map} \rightarrow \text{F}_{\text{t}}\text{-}\textbf{Map}$ . Combining this with the fact that $\hat {\hom }(i, -)$ lifts to $\text{F}_{\text{t}}\text{-}\textbf{Map}$ (which follows by similar arguments to those used in the proof of Lemma 6.6), $\hat {\hom }(i, t_{(-)})$ lifts to a functor

$$ \begin{align*} \widehat{\hom}(i, t_{(-)}) \colon \text{F}\text{-}\textbf{Map} \to \text{F}_{\text{t}}\text{-}\textbf{Map} \rlap{,} \end{align*} $$

which we can apply to f to obtain a uniform trivial fibration $\hat {\hom }(i, t_f)$ .

By part (M2) of Definition 6.1, for every object $X \in \mathbb {E}$ , the map $\emptyset \rightarrow X$ is in $\mathcal {M}$ . Using this, we obtain a morphism H as the canonical filler in

It is straightforward to verify that this H is actually an homotopy from $r_f \circ t_f$ to $1_{P f}$ . This shows that $t_f$ is a strong deformation retract of $r_f$ .

We have given the action of the desired lift $r \colon \text{F}\text{-}\textbf{Map} \rightarrow \underline {\text{SDR}} $ on objects. To show that this construction is functorial on f, consider a morphism of $\text{F}\text{-}\textbf{Map} (h, k) \colon f' \rightarrow f$ . Since the factorisation of the diagonal is functorial, we obtain the diagram

The bottom square is a morphism of $\text{F}_{\text{t}}\text{-}\textbf{Map}$ since it is the result of applying the lift of $t_{(-)}$ of (6.6) to the square $(h, k)$ . Let us prove that $(h, P(h, k)) \colon r_{f'} \rightarrow r_{f}$ is a morphism of strong deformation retracts. Looking at the definition of a morphism of homotopy equivalences (in the paragraph before [Reference Gambino and Sattler14, Lemma 8.1]), we observe that the only thing we need to show is that the following diagram commutes:

where the left and right horizontal arrows are the homotopies witnessing that $r_{f'}$ and $r_f$ respectively are strong deformation retracts. For this, we make use of the naturality of the filling operations. Consider the diagrams:

(6.7)

(6.8)

The left square in (6.7) is a morphism in $\mathcal {M}$ since it is Cartesian. The right square of (6.8) is a morphism of $F_t$ -maps since it is the result of applying the lift $\hat {\hom }(i, t_{(-)}) \colon \text{F}\text{-}\textbf{Map} \rightarrow \text{F}_{\text{t}}\text{-}\textbf{Map}$ to the square $(h, k)$ which is, by hypothesis, a morphism of F-maps. Hence, the corresponding lifts cohere.

Since the construction of the maps $\tilde {H}$ and $\tilde {H}'$ is given by a universal property, it is functorial and so the diagram

commutes. Thus, the composition of the bottom horizontal arrows in (6.7) and (6.8) coincide. This makes the filler $L'$ and L in diagrams 6.7 and 6.8 respectively, the same morphism and thus $H \circ P (h, k) = L' = L = P (h , k)^I \circ H'$ , as required.

Proof Recall that the factorisation of the diagonal $\mathcal {P}_I = \langle r, \rho \rangle $ is divided into two functors $r, \rho \colon \mathbb {E}^{\rightarrow } \rightarrow \mathbb {E}^{\rightarrow }$ . By Lemma 6.6, $\rho $ lifts to the category $\text{F}\text{-}\textbf{Map}$ , so it remains to show that $r \colon \mathbb {E}^{\rightarrow } \rightarrow \mathbb {E}^{\rightarrow }$ lifts to a functor $r \colon \text{F}\text{-}\textbf{Map} \rightarrow \text{C}_{\text{t}}\text{-}\textbf{Map}$ . This follows from two observations. First, since the factorisation of the diagonal is stable, r preserves Cartesian squares and thus, by item (M6) in the hypothesis of the theorem, r lifts to $\mathcal {M}$ (considered as a category of arrows), $r \colon \mathbb {E}^{\rightarrow } \to \mathcal {M}$ . Secondly, consider the unit of the orthogonality adjunction of 3.1 and note that since $(C, F_t)$ is algebraically-free on $\mathcal {M}$ , we obtain a morphism in the slice over $\mathbb {E}^{\rightarrow }$ ,

$$ \begin{align*} \eta_{\mathcal{M}} \colon \mathcal{M} \to \text{C}\text{-}\textbf{Map} \rlap{.} \end{align*} $$

We can compose these last two lifts to obtain $r \colon \mathbb {E}^{\rightarrow } \rightarrow \text{C}\text{-}\textbf{Map}$ .

Finally, we can combine the lifts $r \colon \text{F}\text{-}\textbf{Map} \rightarrow \underline {\text{SDR}}$ from Lemma 6.7 with $r \colon \mathbb {E}^{\rightarrow } \rightarrow \text{C}\text{-}\textbf{Map}$ and apply [Reference Gambino and Sattler14, Proposition 8.5] in order to obtain the desired lift of r,

Proof The result follows from [Reference Gambino and Sattler14, Theorem 8.8] and Proposition 6.8.

Proof We need to verify conditions (M1)–(M6) from the definition of type-theoretic suitable topos. By elementary properties of monomorphisms, it is clear that (M1)–(M3) hold. Condition (M4) follows because in the topos $\mathbb {E}$ the pushout-product construction $\delta ^k \hat {\times } j$ for a given monomorphism $j \colon A \rightarrow B$ , computes the join (or union) of the subobjects $\delta ^k \times B \colon B \rightarrow I \times B$ and $I \times j \colon I \times A \rightarrow I \times B$ , which is again a subobject of $I \times B$ and in particular a monomorphism. The same arguments applies for condition (M5). Finally condition (M6) follows since for any map $f \colon X \rightarrow Y$ , the morphism $r_f \colon X \rightarrow Pf$ is the section of the target map $t_f \colon Pf \rightarrow X$ and in particular, it is a monomorphism.

Proof If we apply $(j \hat {\times } -) \colon \mathbb {E}^{\rightarrow } \rightarrow \mathbb {E}^{\rightarrow }$ to the k-squash square of $i \colon A \rightarrowtail B$ , using that the pushout-product is symmetric and associative, we get the following square:

where we only need to verify that the top horizontal arrow $\Theta $ is the squash morphism, that is, we need to verify that $\Theta = \textbf{sq}_k(j \hat {\times } i) \colon \text{dom}(\delta ^k \hat {\times } (j \hat {\times } i)) \rightarrow D \times B$ , but this follows since the diagram commutes.

Proof This follows intuitively by noticing that the structure of a normal uniform fibration does not add any new lifting problems to that of a uniform fibrations; this is because the only new vertical arrows we are adding are identities and every morphism has a unique lift against them. Concretely, if $(f, \phi ) \in \underline {\text{NrmUniFib}}$ and if $(f, \theta ) \in \underline {\text{UniFib}}$ , then both lifting structures $\phi $ and $\theta $ produce lifts against the exactly same squares, the difference is that $\phi $ may have additional coherence properties.

Proof Let us first assume that $(f, \theta )$ is a normal uniform fibration. It is easy to see that item (2) holds; for this consider the diagram:

it is clear that the lifts cohere because the left square is by definition a morphism in (the image of) $\mathcal {I}^{\textbf{n}}_{\hat {\times }} \rightarrow \mathbb {E}^{\rightarrow }$ .

It is also easy to see that (2) implies (1); this follows since the uniform fibration structure $\theta $ already provides lifts against all lifting problems coming form $\mathcal {I}^{\textbf{n}}_{\hat {\times }}$ ; moreover, the lifts will also cohere with all the squares coming from $u_{\otimes } \colon \mathcal {I}_{\hat {\times }} \rightarrow \mathbb {E}^{\rightarrow }$ . So we only need to verify that it coheres with the squash squares, but these squares are precisely those as in the hypothesis of item (2).

It is clear that (3) implies (2). For the converse let us first observe, using that colimits in $\mathbb {E}$ are universal, that any monomorphism $i \colon A \rightarrowtail B$ is the colimit over the generalised elements with domain on the dense subcategory $\mathbb {A}$ used to define the category of arrows $\mathcal {I}$ (see Proposition 6.4); that is,

$$ \begin{align*} i \cong \text{colim}_{\substack{x\colon \mathscr{A} \rightarrow B \\ \mathscr{A} \in \mathbb{A}}} x^*(i) \rlap{,} \end{align*} $$

where for each $x \colon \mathscr {A} \rightarrow B$ we denote by $x^*(i)$ the pullback of i along x. Now, since $\delta ^k \hat {\times } - \colon \mathbb {E}^{\rightarrow } \rightarrow \mathbb {E}^{\rightarrow }$ is cocontinuous, we have that:

$$ \begin{align*} \text{colim}_{\substack{x\colon \mathscr{A} \rightarrow B \\ \mathscr{A} \in \mathbb{A}}} (\delta^k \hat{\times} (x^*(i))) \cong \delta^k \hat{\times} \text{colim}_{\substack{x\colon \mathscr{A} \rightarrow B \\ \mathscr{A} \in \mathbb{A}}} x^*(i) \cong \delta^k \hat{\times} i. \end{align*} $$

Let us suppose that (2) holds, and that we have a diagram as in item (3). Then for each generalised element $x \colon \mathscr {A} \rightarrow B$ with $\mathscr {A} \in \mathbb {A}$ , we have a square:

where the left square is the colimit inclusion corresponding to $x \colon \mathscr {A} \rightarrow B$ . The commutation of the respective triangle is obtained by the universal property of the colimit.

Finally, if the square on the right factors through a squash square

then (by naturality) the outer square also factors through a squash square and thus the lift $\theta _{x^*(i)}$ is degenerate with $a^* \iota _x $ as lifting degeneracy. This implies by the uniqueness of the universal property, that also $\theta _i$ is degenerate with $a^*$ as lifting degeneracy.

Proof Let $(g, r, h) \in \underline {\text{SDR}}$ which we assume to be $0$ -oriented (the other case being analogous). We have to define $\Psi (g,r,h) := (g, \Psi h)$ with $\Psi h$ a left $\underline {\text{NrmTrivCof}}$ -map structure for g. To do this, let us consider a normal uniform fibration $(f, \phi )$ and a square $(a, b) \colon g \rightarrow f$ for which we will construct a lift $\Psi h_f \colon B \rightarrow X$ as shown:

We first consider the lift $H \colon I \times B \rightarrow X$ , in the following square (which commutes because the deformation retraction is $0$ -oriented), produced by the normal uniform fibration structure of f:

We define $\Psi h_f \colon = H \cdot (\delta ^1 \times B)$ . That is, the lift $\Psi h_f$ is defined to be H on restricted to the top of the cylinder $I \times B$ . The verification that $f \circ \Psi h_f = b$ is straightforward.

We now need to check that $\Psi h_f \cdot g = a$ ; for this we first observe the following diagram:

Here, the lift $H_0$ is also defined by the uniform fibration structure of f, and moreover the triangle created by the lifts commute, since the square on the left is a morphism of left $\underline {\text{UniFib}}$ -maps.

We use that $rg = 1_a$ and the strength of the homotopy retraction tuple $(g, r, h)$ , to replace the horizontal arrows in the previous diagram in order to obtain the following one:

where the lifts cohere by Proposition 7.4 using the squash square of $\bot _A \colon \emptyset \rightarrow A$ . With this in place,

$$ \begin{align*} \begin{array}{llllll} && \Psi h_f \cdot g &= H \cdot (\delta^1 \times B) \cdot g && (\text{by defn of } \Phi h_f) \\ && &= H \cdot (I \times g) \cdot (\delta^1 \times A) && (\text{by naturality of } \delta^1 \times -) \\ && &= H_0 \cdot (\delta^1 \times A) && (\text{by construction of } H_0) \\ && &= a \cdot \rho_1 \cdot (\delta^1 \times A) && (\text{by normality of } (f, \phi)) \\ & & & = a \rlap{,} \end{array} \end{align*} $$

as required.

Proof We construct the desired lift as the following composite:

where the lift in the leftmost square is the forgetful functor, that on the middle square comes from Lemma 6.7 and the lift in the rightmost square is the one from Proposition 7.6.

Proof Since the forgetful functor $\underline {\text{NrmUniFib}} \rightarrow \underline {\text{UniFib}}$ is fully faithful (Lemma 7.3), and using that right orthogonal categories are closed under pullbacks, it is sufficient to prove that given $(f, \psi )$ a normal uniform fibration, the uniform fibration structure $\rho \psi $ of $\hat {\hom }(i, f)$ , described in the foregoing discussion, is also normal.

By Proposition 7.4, we need to show that, for a generating monomorphism $j \colon A \rightarrowtail B$ , the lifts in the diagram on the left of the following figure cohere:

by transposing the whole diagram along $ (i \hat {\times } - ) \dashv \hat {\hom }(i, -)$ and using the symmetry and associativity of the pushout-product, we obtain the lifting problem as on the right of the previous diagram, for which we need to show that the lifts cohere. The lift $\rho \theta _{j}$ on the left (on either diagram) is, by construction, the lift obtained from the uniform fibration structure $\rho \theta $ on $\hat {\hom }(i, f)$ . The result follows by applying Lemma 7.1.

Proof This follows by applying Lemma 8.1 to lift the functor $r \colon \mathbb {E}^{\rightarrow } \rightarrow \mathbb {E}^{\rightarrow }$ and by applying Lemma 6.6 and Lemma 8.3 to lift the functor $\rho \colon \mathbb {E}^{\rightarrow } \rightarrow \mathbb {E}^{\rightarrow }$ .

Proof To show item $(1)$ , let us first consider the following cube:

Here, the square on the top is the pullback of i along f. It is straightforward to verify that all squares pointing from left to right are Cartesian, and note that the squares on the left and right are the outer squares used for defining the pushout-products $\delta ^k \hat {\times } (f^*i)$ and $\delta ^k \hat {\times } i$ respectively. All of this implies that there is a comparison map $\delta ^k \hat {\times } (f^*i) \rightarrow (I \times f)^*(\delta ^k \hat {\times } i)$ , which is an isomorphism because colimits in $\mathbb {E}$ are universal. Item $(2)$ follows directly form item $(1)$ .

Proof The proof uses once again the fact that colimits in $\mathbb {E}$ are universal. Let us compute the pullback of $\delta ^k \hat {\times } i$ along $\delta ^{1-k} \times B$ . By universality of colimits, this is the same as pulling back the diagram defining $B +_A (I \times A)$ and then calculating the colimit.

We can observe in the following picture, the result of first pulling back the defining diagram of $B +_A (I \times A)$ which appears as the upper span of the right-most square on the following cube:

Let us notice that the pullback of $\delta ^k \times B$ (respectively $\delta ^k \times A$ ) along $\delta ^{1-k} \times B$ (respectively $\delta ^{1-k} \times A$ ) is empty since the interval has disjoint endpoints. We conclude that the colimit of the upper span of the left-most square on the cube must be equal to A and moreover, the universal arrow down to B has to be $i \colon A \rightarrow B$ .

Proof Let us denote by $\lambda ^1$ and $\lambda ^2$ , respectively, the canonical trivial uniform cofibration structure on $\delta ^k \hat {\times } (f^*i)$ and the one obtained by applying the functorial Frobenius structure. In order to prove they are the same, let us consider $g \colon Z \rightarrow Y$ a uniform fibration and a square $(a, b) \colon \delta ^k \hat {\times } (f^*i) \rightarrow g$ . Without loss of generality, let us denote by $\lambda ^1$ and $\lambda ^2$ the two fillers of this square given by the uniform trivial cofibration structure with the same name. We have to show that $\lambda ^1 = \lambda ^2$ . If we go over the proof of [Reference Gambino and Sattler14, Proposition 8.8], just before the conclusion, a retract diagram is used to transfer the structure of a trivial cofibration to the desired morphism (since trivial cofibrations are closed under retracts). In our situation, this retract diagram is given by the two left-most squares shown below:

where the left-most square is $\theta ^k \hat {\times } \delta ^k \hat {\times } (f^*i)$ . Notice that $\delta ^k \hat {\times } \delta ^k \hat {\times } (f^*i)$ has a canonical trivial cofibration structure and thus, the square $\delta ^k \hat {\times } \delta ^k \hat {\times } (f^*i) \rightarrow f$ has a lift which we denote by $\lambda $ . By definition, the lift $\lambda ^2$ is equal to $\lambda \cdot t$ where t is the horizontal arrow on the lower left part of the diagram.

On the other hand, the lift of the outer square of the previous diagram is $\lambda ^1$ . Thus if we want to show that $\lambda ^1 = \lambda ^2$ it is sufficient to show that the square $\theta ^k \hat {\times } \delta ^k \hat {\times }(f^*i)$ is a morphism of trivial uniform cofibrations. To show this, we use that the pushout-product is symmetric and associative, and thus $\theta ^k \hat {\times } \delta ^k \hat {\times }(f^*i) \cong \delta ^k \hat {\times } \theta ^k \hat {\times } (f^*i)$ . From this, we see that the square is a morphism of trivial uniform cofibrations provided the square $\theta ^k \hat {\times } (f^*i)$ is a morphism of generating monomorphisms, i.e., if it is Cartesian. But this is precisely the statement of Lemma 8.6.

Proof Object-wise, this follows directly from [Reference Gambino and Sattler14, Theorem 8.8]. To see this, we notice that there are no more objects in the image of the category of arrows $\mathcal {I}^{\textbf{n}}_{\hat {\times }}$ than in the image of $\mathcal {I}_{\hat {\times }}$ ; thus we can apply the functorial Frobenius structure for uniform fibrations. Then we use the functor $\underline {\text{TrivCof}} \rightarrow \underline {\text{NrmTrivCof}}$ , obtain by functoriality of the left orthogonal functor applied to the forgetful functor $\underline {\text{NrmUniFib}} \rightarrow \underline {\text{UniFib}}$ .

For the morphism case, we first notice that the only morphisms in $\mathcal {I}^{\textbf{n}}_{\hat {\times }}$ that we need to consider are the squash squares. Thus let us consider a cospan of squares as in the following diagram:

such that the vertical square is the squash square of a generating monomorphism $i \colon A \rightarrowtail B$ and the horizontal square is a morphism of uniform fibrations $(m , \epsilon \times B) \colon f' \rightarrow f$ . We need to verify that pulling back the squash square along the morphism of uniform fibrations is a morphism of normal trivial cofibrations.

The first thing we do is to split this cospan of squares into two, by factoring through the pullback square of f along $\epsilon \times B$ . That is we obtain the following diagrams:

where the dotted arrow $m^* \colon X' \rightarrow (I \times X)$ is obtain by universal property. Notice that composing the two cospans of squares along their common face produces the original one. Notice also that the two horizontal squares are morphisms of uniform fibrations.

Let us focus first on the cospan of the right. The identity morphism $1 \colon (\delta ^k \hat {\times } i) \rightarrow (\delta ^k \hat {\times } i)$ is a morphism of trivial uniform cofibrations, thus if we pull-back this along the morphism of uniform fibrations $(f', I \times f) \colon m^* \rightarrow 1_{\delta ^k \hat {\times } i}$ we obtain a morphism of trivial uniform cofibrations by [Reference Gambino and Sattler14, Theorem 8.8] to which we can apply the functor $\underline {\text{TrivCof}} \rightarrow \underline {\text{NrmTrivCof}}$ to obtain a morphism of normal trivial cofibrations.

With this we have reduced the situation to the cospan of squares on the left of the previous diagram. Using item $(2)$ of Lemma 8.5 we see that the pullback of the squash square of $i \colon A \rightarrowtail B$ along the square $(I \times f, f) \colon \epsilon \times X \rightarrow \epsilon \times B$ is the squash square of $f^*i \colon f^*A \rightarrowtail X$ . This square is a morphism in $\mathcal {I}^{\textbf{n}}_{\hat {\times }}$ provided that the canonical trivial normal cofibration structure of $\delta ^k \hat {\times } (f^*i)$ is the same as that obtained from the functorial Frobenius structure, but this follows from Lemma 8.7.

Proof Using the lift of Proposition 8.8 and the forgetful functor $\underline {\text{NrmUniFib}} \rightarrow \underline {\text{UniFib}}$ , we find a lift of the pullback functor as one shown:

The fact that we can extend this structure from $\mathcal {I}^{\textbf{n}}_{\hat {\times }}$ to the whole category $\underline {\text{NrmTrivCofi}}$ follows from [Reference Gambino and Sattler14, Proposition 6.8].

Proof The claim follows from Proposition 8.4 and Proposition 8.9.