Proof. The category $\oint _{\mathcal{A}}X$ is isomorphic to the coproduct of its maximal connected subcategories:

\begin{equation*}\oint _{\mathcal{A}}X\cong \coprod _{s\in \mathcal{S}}\mathcal{C}_s\end{equation*}

Note that the uniqueness of the predegenerate cell of $a\in X_P$ in the definition of an EZ-presheaf corresponds to the existence of a weakly terminal object in the category $\mathcal{C}_s$ containing $a$ . The uniqueness of the degenerating map means precisely that this weakly final object is, in fact, terminal in a strict sense.

Proof. Take the morphism $R\xrightarrow {\mu } P$ corresponding to the cell $a$ via Yoneda embedding. Using the assumption that $\mathcal{A}$ is a Reedy category, we factor it as $R\to Q\xrightarrow {\phi } P$ , where $\phi$ is a coface map. Since $a$ is not in $\partial {\unicode{x3088}_P}$ , $\phi =1_P$ and hence $\mu$ must be a codegeneracy map.

Proof. Let $ {\unicode{x3088}_P}$ be a representable presheaf and $x\in ( {\unicode{x3088}_P})_Q$ . By the Yoneda lemma, we can equivalently view $x$ as a morphism $x: Q\to P$ . We can factor $x=\delta \circ \sigma$ uniquely, where $\delta$ is a coface map and $\sigma$ is a codegeneracy map. The cell of $ {\unicode{x3088}_P}$ corresponding to the morphism $\delta$ is the only nondegenerate cell with $x$ as its image by some degeneracy map. This degeneracy map can only be $\sigma ^*$ .

Proof. Since colimits are pointwise in presheaves, for every $Q$ , each cell in $\widetilde {X}_Q\setminus X_Q$ is a unique degeneracy of the image of the main cell in $ {\unicode{x3088}_P}$ via $\hat {\phi }$ and conversely, if some cell $a\in \widetilde {X}_Q$ is a degeneracy of the image of the main cell in $ {\unicode{x3088}_P}$ , then $a\in \widetilde {X}_Q\setminus X_Q$ .

Let $a\in \widetilde {X}_Q$ for some $Q$ . If $a\in X_Q$ , then there is a unique cell $a'\in X_Q^{nd}$ which degenerates to $a$ , since $X$ is an EZ-presheaf. By what was written above, it follows that $a'$ is also nondegenerate in $\widetilde {X}$ and no other nondegenerate cell of $\widetilde {X}$ degenerates to $a$ .

Similarly, if $a\in \widetilde {X}\setminus X$ , then $a$ is a degeneracy of $\hat {\phi }$ . By the same reasoning as in Lemma3.13, $a$ is not a degeneracy of any nondegenerate cell distinct from $\hat {\phi }$ .

More compactly, we have shown that $\oint _{\mathcal{A}}\widetilde {X}$ is obtained from $\oint _{\mathcal{A}} X$ just by adding one new connected component, corresponding to degeneracies of the cell $\hat {\phi }$ .

Proof. We will only show that the three conditions listed above imply that $g$ is a cellular monomorphism (the other implication, not using the degeneracy-forking condition, is left to the reader).

Using transfinite induction, it suffices to show that if $\alpha$ is a nondegenerate cell in $W_P\setminus Z_P$ with boundary contained in $Z$ , then the map $g'$ shown on the diagram is a monomorphism that satisfies the same conditions $\textit {(1)-(3)}$ as the map $g$ .

First, we will show that $g'$ is a monomorphism.

Suppose that there are two distinct cells $e, e'\in \big (Z\amalg _{\partial {{\unicode{x3088}_P}}} {{\unicode{x3088}_P}}\big )_Q$ such that $g'(e)=g'(e')$ . Without loss of generality, $e\not \in Z_Q$ , so the previous lemma implies that it must be a degeneracy of a top cell of $ {\unicode{x3088}_P}$ .

If $e'\not \in Z_Q$ , then $e'$ is also a degeneracy of a top cell of $ {\unicode{x3088}_P}$ . The top cell of $ {\unicode{x3088}_P}$ is sent to $\alpha$ , which is assumed to be nondegenerate in $W$ , hence by $\textit {(1)}$ distinct degeneracy operators acting on $\alpha$ give different cells, so $e$ and $e'$ must be mapped by $g$ to different degeneracies of $\alpha$ .

If $e'\in Z_Q$ , then the pair of cells $e',\alpha$ violates the condition $it{(2)}$ .

Conditions $\textit {(1)}$ and $\textit {(3)}$ for $g'$ are obvious, since the image of $g'$ contains the image of $g$ .

To verify the condition $\textit {(2)}$ , we use the assumption that $\mathcal{A}$ is degeneracy-forking. Suppose that there exists $d\in \big (Z\amalg _{\partial {{\unicode{x3088}_P}}} {{\unicode{x3088}_P}}\big )_Q$ such that one can find $e\in \big (W\setminus Z\amalg _{\partial {{\unicode{x3088}_P}}} {{\unicode{x3088}_P}}\big )_R$ which degenerates to $d$ . Since the condition $\textit {(2)}$ was satisfied for $g$ , we note that $d$ must be a degeneracy of a top cell of $ {\unicode{x3088}_P}$ .

We can assume that $e$ is nondegenerate in $W$ and that $d$ is nondegenerate in $Z\amalg _{\partial {\unicode{x3088}_P}} {\unicode{x3088}_P}$ . Explicitly, let $e'$ be the predegenerate cell of $e$ (unique by condition $\textit {(3)}$ ) in a presheaf $W$ and let $d'$ be some predegenerate cell of $d$ in a presheaf $Z$ (obviously $e'$ is not an image of any cell in $Z$ , since otherwise, $e$ would be an image of a cell in $Z$ ). We note that $e'$ degenerates to $e$ which degenerates to $d$ and also $d'$ degenerates to $d$ . Let $d''$ be the predegenerate cell of $d''$ in $Y$ . Both $e', d''$ are nondegenerate cells of an EZ-presheaf $W$ which degenerate to the same cell $d$ , and since $e'\not \in Z_Q$ , they must be equal by $\textit {(3)}$ . In particular, we can assume that $d$ is a top cell of $P$ .

Let $\sigma : R\to Q\in \textrm {Mor}(\mathcal{A})$ be the codegeneracy map corresponding to the degeneracy operator sending $e$ to $d$ . Let the pair $\sigma ',\sigma '':S\to R$ of distinct codegeneracy morphisms satisfy $\sigma \circ \sigma '=\sigma \circ \sigma ''$ . Then $\sigma '^*(d)\neq \sigma ''^*(d)$ since there are no identifications between degeneracies of the top cell of $P$ in the pushout $Z\amalg _{\partial {\unicode{x3088}_P}} {\unicode{x3088}_P}$ . On the other hand, they must be equal, since $d=\sigma ^*(e)$ . A contradiction.

Proof. Suppose $X\overset {f}{\hookrightarrow }Y$ is a monomorphism between EZ-presheaves, which is not cellular. This means that some cell $e\in \big (Y\setminus \textrm {Im}(f)\big )_Q$ has a degeneracy (by the operator $\sigma ^*$ ) which is an image of a cell $d\in X_R$ (this is the negation of the condition $\textit {(2)}$ from the previous lemma, conditions $\textit {(1)}$ and $\textit {(3)}$ are trivially satisfied, since $X$ and $Y$ are EZ-presheaves).

Since $X$ and $Y$ are EZ-presheaves, we can assume that $e$ is nondegenerate in $Y$ and $d$ is nondegenerate in $X$ . Explicitly, let $e'$ be the predegenerate cell of $e$ in a presheaf $Y$ and let $d'$ be the unique predegenerate cell of $d$ in a presheaf $X$ (obviously $e'$ is not an image of any cell in $X$ , since otherwise, $e$ would be an image of a cell in $X$ ). It suffices to observe that $e'$ degenerates to $f(d')$ . Indeed, $e'$ degenerates to $e$ which degenerates to $f(d)$ and also $f(d')$ degenerates to $f(d)$ . Let $d'$ be the predegenerate cell of $f(d')$ in $Y$ . Both $e', d'$ are nondegenerate cells of an EZ-presheaf $Y$ which degenerate to the same cell $f(d)$ , hence they must be equal. In other words, $e'$ is a predegenerate cell of $f(d')$ in a presheaf $Y$ .

Let $\sigma : R\to Q\in \textrm {Mor}(\mathcal{A})$ be the codegeneracy map corresponding to the degeneracy operator sending $e$ to $f(d)$ . Let the pair $\sigma ',\sigma '':l\to R$ of distinct codegeneracy morphisms satisfy $\sigma \circ \sigma '=\sigma \circ \sigma ''$ . Then $\sigma '^*(f(d))\neq \sigma ''^*(f(d))$ since $d$ is a nondegenerate cell in EZ-presheaf and $f$ is a monomorphism. On the other hand, they need to be equal, since $f(d)=\sigma ^*(e)$ . A contradiction.

Proof. Initial presheaf is both an EZ-presheaf and a monomorphism $\emptyset \to \emptyset$ is cellular. Since $\emptyset \hookrightarrow X$ is mono, the previous result implies that it is a cellular monomorphism. As a result, $\emptyset \to X$ is a cellular monomorphism.

Proof. Let $X\xrightarrow {f}Y$ be a cellular monomorphism and consider the following diagram in which compositions in both rows are identity morphisms:

Morphism $g$ is a monomorphism, being a retract of a monomorphism. Suppose that $g$ is not a cellular monomorphism, so at least one of the conditions specified in Lemma3.15 does not hold. We show that each of those possibilities leads to a contradiction:

1. (1) there is a nondegenerate cell $e\in W_P\setminus Z_P$ such that $\sigma ^*(e)=\sigma '^*(e)$ for some pair of distinct codegeneracy maps $\sigma , \sigma '$ . Since this is not the case for $X$ and $Y$ , this means that $i'(e)\in Y$ is degenerate. We pick its predegenerate cell $e'$ (we note that it is of a different shape than $e$ ). Then $r'(e')$ is a predegenerate cell of $e$ in $W$ , a contradiction (more generally, this reasoning shows that the image of a nondegenerate cell in $W$ is nondegenerate in $Y$ , which we will use again in the point $(3)$ ).

2. (2) there is a pair of cells $e\in W_P, e'\in W_Q$ such that $e$ degenerates to $e'$ and $e'\in Z_Q$ , $e\not \in Z_P$ . Since $f$ is a cellular monomorphism, $i'(e)\in X_P$ (otherwise $f$ would violate a condition $(2)$ ). Then $e=r'\circ i'(e)=r'\circ f\circ i'(e)=g\circ r\circ i'(e)\in Z$ , a contradiction.

3. (3) There are distinct nondegenerate cells $e\in W_P, e'\in W_Q\setminus Z_Q$ which degenerate to the same cell. The cell $i'(e')$ is not in the image of $f$ : if $f(d)=i'(e')$ , then $g(r(d))=r'(f(d))=r'(i'(e'))=e'$ . It follows that $i'(e)$ and $i'(e')$ violate condition $(2)$ for the cellular monomorphism $f$ .

Proof. The first claim follows from the small object argument, see Riehl (Reference Riehl2014).

For the second one, note that if $X$ is a colimit of a sequence of monomorphisms between EZ-presheaves, then $X$ is an EZ-presheaf itself. Combine that observation with the Theorem3.14 and the fact that $\emptyset$ is an EZ-presheaf.

The last is a combination of the second claim with Corollary3.18.

Proof. By Corollary3.21, $X$ is an EZ-presheaf if and only if $\emptyset \to X$ is a cellular monomorphism, we can apply Lemma3.19. Alternatively, one can prove that directly from the definition of an EZ-presheaf.

Proof. Cavallo and Sattler (Reference Cavallo and Sattler2022), Lemma5.29.

Proof. Let $(a,b)\in \big (X\times Y\big )_P$ and $(a',b')\in \big (X\times Y\big )_Q, (a'',b'')\in \big (X\times Y\big )_R$ be two nondegenerate cells of $X\times Y$ which degenerate to $(a,b)$ :

\begin{equation*}\mu ^*(a',b')=(a,b)=\rho ^*(a'',b'')\end{equation*}

Both cells $a'$ and $ a''$ degenerate to $a$ . Since $X$ is an EZ-presheaf, there is $a'''\in X_S$ which degenerates to $a'$ and also to $a''$ . This must factor through the cell which has the shape of the pushout of $\rho$ and $\mu$ , call it $\hat {a}\in X_{Q\amalg _P R}$ . Cell $\hat {a}$ degenerates to $a$ via an induced pushout map, which we denote by $\eta$ .

Similarly, we obtain $\hat {b}\in Y_{Q\amalg _P R}$ which degenerates to $b$ via $\eta$ . But then $(\hat {a},\hat {b})\in \big (X\times Y\big )_{Q\amalg _P R}$ degenerates to both $(a',b')$ and $(a,b)$ and since those are nondegenerate, must be equal to both of them.

Now suppose that $(a,b)\in \big (X\times Y\big )_P^{nd}$ and $\mu , \rho$ are codegeneracy maps s. t. $\mu ^*(a,b)=\rho ^*(a,b)=(a',b')\in \big (X\times Y\big )_Q$ . It follows that $\mu ^*a=\rho ^*a=a'$ and since $X$ is an EZ-presheaf, $a$ must be a degeneracy of some cell $a''$ via $\kappa$ s. t. $\mu \circ \kappa =\rho \circ \mu$ . Cell $a''$ must degenerate to some cell $\hat {a}\in X_S$ , where $S$ is a pushout of $\mu$ and $\rho$ . It degenerates to $a$ via the canonical pushout morphism $\eta$ . Note that $\hat {a}$ must be distinct from $a$ since the pushout of two distinct morphisms is not an identity map (hence it is a proper codegeneracy map and therefore domain is not equal to the codomain).

Similarly, we obtain a cell $\hat {b}\in Y_S$ which degenerates to $b$ via $\eta$ , but then $\eta ^*(\hat {a},\hat {b})=(a,b)$ . Since $(a,b)$ is a nondegenerate cell, it follows that $(\hat {a},\hat {b})=(a,b)$ and $\eta =\mathrm{Id}_S$ . Since the pushout of two degeneracy maps is an identity only if those maps are equal (since otherwise it would be a morphism strictly lowering the degree), it follows that $\mu =\rho$ .

Proof. Let $c$ be the unique nondegenerate cell of $X$ which degenerates to $a$ and $b$ , that is $\sigma _a^* c=a$ , $\sigma _b^*c=b$ . Then $\sigma ^*\sigma _a^*c=\sigma ^*\sigma _b^*c$ , so by the uniqueness of degeneracy operator $\sigma _a\circ \sigma =\sigma _b\circ \sigma$ . Since $\sigma$ is epi, $\sigma _a=\sigma _b$ and consequently $a=\sigma _a^*c=\sigma _b^*c=b$ .

Proof. $(\implies )$ Fix a comparing map $g: {\unicode{x3088}_Q}\amalg _{ {\unicode{x3088}_P}} {\unicode{x3088}_R}\to {\unicode{x3088}_S}$ .

First, we will show the existence of a dashed arrow. Let $c\in X_T$ be a common predegenerate cell of $f(\alpha _Q)$ and $f(\alpha _R)$ . By the universal property of a pushout, there exists a unique codegeneracy map $\kappa : S\to T$ . Then $\kappa ^*c: {\unicode{x3088}_S} \to X$ is the desired map.

It remains to justify the uniqueness of such a map. Let $\sigma ': R\to S$ be a pushout of $\sigma$ along $\tau$ and suppose that $f_1, f_2: {\unicode{x3088}_S}\to X$ satisfy $f=f_1\circ g=f_2\circ g$ . Then $\sigma '^* f_1=f(\alpha _r)=\sigma '^* f_2$ . Now it suffices to apply Lemma3.29.

$({\impliedby})$ Suppose that $X\not \in \widehat {\mathcal{A}}_{EZ}$ . This means that at least one of the following conditions holds:

1. (1) there exists a pair of distinct cells $a\in X_Q^{nd}$ , $b\in X_R^{nd}$ and a pair of codegeneracy maps $\sigma ,\tau$ such that $\sigma ^*a=\tau ^*b\in X_P$

2. (2) there exists a cell $a\in X_Q^{nd}$ and a pair of distinct codegeneracy maps $\sigma ,\tau$ such that $\sigma ^*a=$ $\tau ^*b\in X_P$

In the first of those cases, $X$ is not orthogonal to a comparing map $ {\unicode{x3088}_Q}\amalg _{ {\unicode{x3088}_P}} {\unicode{x3088}_R}\to {\unicode{x3088}_S}$ : if it was, then the image of $\alpha _S$ in $X$ would be a common predegenerate cell of $a$ and $b$ (which does not exist for a pair of distinct nondegenerate cells).

In the second case, we reason as follows: since $\sigma$ and $\tau$ are distinct, their pushouts are not identities, in particular $d(S)\lt d(Q)$ . If $X$ is orthogonal to a comparing map $ {\unicode{x3088}_Q}\amalg _{ {\unicode{x3088}_P}} {\unicode{x3088}_Q}\to {\unicode{x3088}_S}$ , then the image of $\alpha _S$ in $X$ is a predegenerate cell of $a$ which has a strictly smaller dimension than $a$ , contradicting the fact that $a$ is a nondegenerate cell.

Proof. Adámek and Rosický (Reference Adámek and Rosický1994), Theorem 1.39.

Proof. The morphism $f$ is a cellular monomorphism, being a monomorphism between EZ-presheaves. Since cellular monomorphisms are closed under taking pushouts, $\tilde {f}: Z\hookrightarrow Y\amalg _X Z$ is also a cellular monomorphism. Since $Z$ is an EZ-presheaf, $\emptyset \hookrightarrow Z$ is a cellular monomorphism. It follows that $\emptyset \hookrightarrow Y\amalg _X Z$ is cellular, so $Y\amalg _X Z$ is an EZ-presheaf.

Proof. Direct check.

Proof. This follows from Adámek and Rosický (Reference Adámek and Rosický1994), p. 31, and the Theorem3.31 by observing that both domains and codomains of comparing maps are finitely presentable.

Proof. Suppose that $\pi _0\circ \oint _{\mathcal{A}}f:\pi _0\circ \oint _{\mathcal{A}}X\to \pi _0\circ \oint _{\mathcal{A}}Y$ is not a surjection.

The functor $\pi _0\circ \oint _{\mathcal{A}}$ is a left adjoint (being a composition of three left adjoints), in particular, it preserves epimorphisms. It follows that $f$ is not an epimorphism.

Conversely, assume that $\pi _0\circ \oint _{\mathcal{A}}f:\pi _0\circ \oint _{\mathcal{A}}X\to \pi _0\circ \oint _{\mathcal{A}}Y$ is a surjection and let $k,h:Y\to Z$ be a pair of morphisms between EZ-presheaves such that $k\circ f=h\circ f$ .

It is enough to check that $k$ and $h$ agree on each cell of $Y$ . Fix $\alpha \in Y$ . By the assumption, there exists a codegeneracy map $\sigma$ and a cell $\beta \in X$ such that $f(\beta )=\sigma ^*\alpha$ , so $\sigma ^*k(\alpha )=k\circ f(\beta )=h\circ f(\beta )=\sigma ^*h(\alpha )$ . Applying Lemma3.29, we finish the proof.

Proof. From the assumptions follows that $f$ is a cellular monomorphism such that for every nondegenerate $\alpha \in Y$ there exists $\beta \in X$ and a codegeneracy map $\sigma$ such that $f(\beta )=\sigma ^*\alpha$ .

It is enough to check that every nondegenerate cell of $Y$ is an image of some cell of $X$ by $f$ , so suppose that $\alpha \in Y_P\setminus X_P$ . This contradicts the point $(3)$ of the Lemma3.15.

Proof. The diagram

where the morphism on the left is a generic comparing map, expressing the fact that $R(X)$ is an EZ-presheaf.

By the standard correspondence of lifting problems under adjunction, the existence of the unique filler in this diagram is equivalent to

which is true by the assumption on $F$ (the arrow on the left is an isomorphism).

Proof. Ara et al. (Reference Ara, Burroni, Guiraud, Malbos, Métayer and Mimram2025), Lemma 1.3.19.

Proof. Ara et al. (Reference Ara, Burroni, Guiraud, Malbos, Métayer and Mimram2025), Lemma 1.3.21.

Proof. We will use Newman’s lemma.

Step 1: Reduce the number of generating reductions.

We observe that every relation in $\leadsto$ decomposes into elementary steps:

\begin{equation*} (x_0, \ldots , x_{i-1}, x_i, \ldots , x_n) \simeq (x_0, \ldots , x_{i-1}, \gamma (x_i), \ldots , x_n), \end{equation*}

where $x_{i-1} \preceq \gamma (x_i)$ .

Step 2: Termination.

This system terminates because reductions strictly decrease the sum of cell dimensions in a sequence.

Step 3: Local confluence.

Consider sequences with reductions at positions $i$ and $j$ . If $x_{i-1} \preceq \gamma (x_i)$ and $x_{j-1} \preceq \gamma (x_j)$ , then the sequence $(x_0', x_1',\ldots , x_{n-1}', x_n')$ with $x_l'=x_l$ for $l\neq i,j$ , $x_i'=\gamma (x_i)$ , $x_j'=\gamma (x_j)$ remains valid in $\mathcal{N}_{\textsf{pOpe}_\iota}(P)_n$ .

For the sake of transparency, we spell out explicitly the only non-obvious case $j=i+1$ :

\begin{align*} &(x_0, \ldots , x_{i-1}, x_i, x_{i+1}, \ldots , x_n) \simeq (x_0, \ldots , x_{i-1}, \gamma (x_i), x_{i+1}, \ldots , x_n), \\ &(x_0, \ldots , x_i, x_{i+1}, x_{i+2}, \ldots , x_n) \simeq (x_0, \ldots , x_i, \gamma (x_{i+1}), x_{i+2}, \ldots , x_n). \end{align*}

We need to show that $(x_0,\ldots , x_{i-1}, \gamma (x_{i}), \gamma (x_{i+1}),x_{i+2},\ldots , x_n)$ is an element of $\mathcal{N}_{\textsf{pOpe}_\iota}(P)_n$ . For this, we note that $x_{i}\preceq \gamma (x_{i+1})$ implies $\gamma (x_{i})\preceq \gamma (x_{i+1})$

Proof. Step 1: Reduction to generators.

By Observation4.16 it suffices to check this for generators, that is monomorphisms of the form $\partial _P:\partial {\unicode{x3088}_P}\to {\unicode{x3088}_P}$ .

Step 2: Reduction to $\mathbf{Set}$ .

In $\textsf{sSet}$ every monomorphism is cellular, so it is enough to show that $\zeta (\partial _P)$ is a monomorphism. Furthermore, since monomorphisms in $\textsf{sSet}$ are pointwise, we only need to show that $\zeta (\partial _P)_n$ is a monomorphism in $\mathbf{Set}$ for every $n\in \mathbb{N}$ .

Step 3: Exploitation of the confluence property.

Let us fix a pair $[x_0,\ldots , x_n], [y_0,\ldots , y_n]\in \zeta (\partial {\unicode{x3088}_P})$ of simplices, which are mapped to the same simplex in $\zeta ( {\unicode{x3088}_P})$ . We can assume that $(x_0,\ldots , x_n)$ is in a normal form (when considered as an element of $ \unicode{x3088}_{\langle x_n\rangle }$ and similarly for $(y_0,\ldots , y_n)$ .

Then $\mathfrak{Y}(\partial _P)(x_0,\ldots , x_n)$ and $\mathfrak{Y}(\partial _P)(y_0,\ldots , y_n)$ have a common reduction. On the other hand, if a chain $(x_0,\ldots , x_n)$ is reduced in $\langle x_n\rangle$ , then it must be also reduced in $P$ , hence $(x_0,\ldots , x_n)=(y_0,\ldots , y_n)$ .

Proof. Verification that formulas given in the statement yield well-defined morphisms of simplicial sets is straightforward.

Colimits in presheaves are computed pointwise, so it suffices to verify the pushout property in $\mathbf{Set}$ after evaluating on a representable. Given a cone:

we construct a unique $f \colon \zeta (P)_n \to X$ . For a reduced chain $(x_0, \ldots , x_n)$ :

Case 1: $[(x_0, \ldots , x_n)] \in \zeta (\langle \gamma ^2 \alpha _P\rangle )$ .

If $[(x_0,\ldots ,x_n)] \in \zeta (\langle \gamma ^2 \alpha _P\rangle )$ , then $f(x_0,\ldots ,x_n)$ is forced to agree with the inclusion map $\zeta (\langle \gamma ^2 \alpha _P\rangle ) \overset {}{\hookrightarrow } X$ .

Furthermore, suppose there exists an element $[(y_0,\ldots ,y_k) \ast (\bullet ,\ldots ,\bullet )]$ in $\zeta (\delta P) \ast \{\bullet \}$ (where $(y_0,\ldots ,y_k)$ is reduced in $y_k$ ) that maps to $[(x_0,\ldots ,x_n)] \in \zeta (P)$ . By confluence, $(y_0,\ldots ,y_k) = (x_0,\ldots ,x_k)$ . Consequently, $[(y_0,\ldots ,y_k) \ast (\bullet ,\ldots ,\bullet )]$ lies in $\zeta (\langle \gamma ^2 \alpha _P\rangle ) \ast \{\bullet \}$ , thus verifying that $f$ defines a map of cones.

Case 2: $[(x_0, \ldots , x_n)] \in \zeta (\delta P) \setminus \zeta (\langle \gamma ^2 \alpha _P\rangle )$ .

If $[(x_0,\ldots ,x_n)] \in \zeta (\delta P) \setminus \zeta (\langle \gamma ^2 \alpha _P\rangle )$ , then $f(x_0,\ldots ,x_n)$ is forced to agree with the composite inclusion $\zeta (\delta P) \hookrightarrow {} \zeta (\delta P) \ast \{\bullet \} \hookrightarrow {} X$ .

Assume there exists an element $[(y_0,\ldots ,y_k) \ast (\bullet ,\ldots ,\bullet )]$ in $\zeta (\delta P) \ast \{\bullet \} \setminus \zeta (\delta P)$ (again with $(y_0,\ldots ,y_k)$ reduced in $y_k$ ) that maps to $[(x_0,\ldots ,x_n)] \in \zeta (P)$ . By confluence, $(y_0,\ldots ,y_k) = (x_0,\ldots ,x_k)$ , and all subsequent terms $x_{k+1},\ldots ,x_n$ must equal $\gamma (\alpha _P)$ or $P$ . However, since $[(x_0,\ldots ,x_n)] \in \zeta (\delta P) \setminus \zeta (\langle \gamma ^2 \alpha _P\rangle )$ , the term $x_{k+1}$ cannot be reduced to $\gamma ^2 (\alpha _P)$ , which yields a contradiction with $x_{k+1}\in |\delta P|$ .

Case 3: $[(x_0, \ldots , x_n)] \in \zeta (P) \setminus \zeta (\delta P)$ .

Chain $(x_0,\ldots ,x_n)$ contains a face $x_i \in \delta P \setminus \gamma ^2 (\alpha _P)$ , and the smallest subsequent term $x_j$ (with $i \lt j$ ) equals $\gamma (\alpha _P)$ or $P$ . We define $f(x_0,\ldots ,x_n)$ to coincide with the value of the inclusion $\zeta (\delta P) \ast \{\bullet \} \hookrightarrow {} X$ applied to $(x_0,\ldots ,x_{j-1},\bullet ,\ldots ,\bullet )$ . An argument analogous to the previous case demonstrates that there is exactly one element in $\zeta (\delta P) \ast \{\bullet \}$ mapping to $(x_0,\ldots ,x_n)$ , ensuring uniqueness.

Proof. Follows directly from the pencil-linearity applied to $a$ and the first globularity axiom (if $\gamma (d)=b$ , then the last face $e$ in the poset $(\delta (b),\lt ^{-})$ satisfies $\gamma (e)=\gamma (b)$ ).