Proof. In order to show that $f\circ _i g$ belongs to $\mathcal {L}(X^{d+e-1},X)$ we need to show that for any $c\in X$ and an arbitrary choice of $d+e-2$ of the $d+e-1$ elements $x_1,x_2,..,x_{d+e-1}\in X$ the remaining of these elements is uniquely determined by the equation

(1) \begin{equation} f(x_1,..,x_{i-1}, g(x_i,..,x_{i+e-1}), x_{i+e},..,x_{d+e-1}) = c. \end{equation}

Let $s$ be an integer such that $1\leq s\leq d+e-1$ . Fix elements $x_1,x_2,..,x_{s-1}, x_{s+1},..,x_{d+e-1}, c \in$ $X$ .

Consider first that case where $s\leq i-1$ or $s\geq i+e$ . Then, setting $y=g(x_i,..,x_{i+e-1})$ , the Equation 1 becomes

(2) \begin{equation} f(x_1,..,x_{i-1},y,x_{i+e},..,x_{d+e-1}) = c. \end{equation}

All entries but the entry $x_s$ in this equation are fixed. Since $f\in \mathcal {L}(X^d,X)$ it follows that there is a unique choice for $x_s\in X$ such that Equation 2 holds, and hence a unique choice for $x_s\in X$ such that Equation 1 holds.

Consider the remaining case where $i\leq s \leq i+e-1$ . Then in particular the elements $x_1,..,x_{i-1}, x_{i+e},..,x_{d+e-1}$ are fixed in $X$ . Since $f\in \mathcal {L}(X^d,X)$ it follows that there is a unique $y\in X$ such that Equation 2 holds. Thus for Equation 1 to hold, we must have

(3) \begin{equation} g(x_i,..,x_{i+e-1}) = y. \end{equation}

In this equation all but $x_s$ have been chosen. Since $g\in \mathcal {L}(X^e,X)$ it follows that there is a unique choice $x_s\in X$ such that Equation 3 holds. In all cases, there is a unique choice of $x_s$ such that Equation 1 holds. This shows that $f\circ _i g$ belongs to $\mathcal {L}(X^{d+e-1},X)$ and completes the proof.

Proof of Theorem 1.2. Let $d\in \mathbb {N}$ . A map $f\in \mathcal {E}(d)=$ $\mathrm {Map}(X^d,X)$ belongs to $\mathcal {L}(d)=\mathcal {L}(X^d,X)$ if and only if the set

\begin{equation*}L = \{(x_1,x_2,..,x_d, f(x_1,x_2,..,x_d))\ |\ x_1,x_2,..,x_d \in X\}\end{equation*}

is a Latin hypercube in $X^{d+1}$ . Equivalently, $f$ belongs to $\mathcal {L}(d)$ if and only for every $c\in X$ and an arbitrary choice of $d-1$ of the $d$ entries $x_1,..,x_d\in X$ the remaining entry is uniquely determined by the equation $f(x_1,x_2,..,x_d)=c$ . The action of $\mathfrak {S}_d$ on $\mathrm {Map}(X^d,X)$ by permuting the $d$ coordinates of $X^d$ clearly preserves the subset $\mathcal {L}(d)$ , and $\mathrm {Id}_X$ belongs to $\mathcal {L}(1)$ . In order to prove Theorem1.2, it remains to show that the sets $\mathcal {L}(d)$ are closed under the operations $-\circ _i-$ . This is done in Lemma2.1 above, and this concludes the proof.

Remark 2.2. We could have very slightly simplified the proof of Lemma 2.1 by observing that thanks to the symmetric group actions on the coordinates of Latin hypercubes (cf. Remark 1.4) it would have been sufficient in the proof of Lemma 2.1 to consider the map $f\circ _{d} g$ and a single $s$ in the distinction into the two cases for $s$ . That is, it would have been sufficient in the last part of the proof of Lemma 2.1 to consider the cases where either $1=s\leq d-1$ or $s=d$ . We will make use of this observation in the proof of the more general Theorem 3.4 below.

Remark 2.3. The proof of Theorem 1.2, as written, involves choices of elements in the set $X$ . In Section  3 below we will rewrite this proof in such a way that it extends to categories with finite products.

Definition 3.1. Let $\mathcal {C}$ be a category with finite products. Let $d$ be a positive integer. We denote by $\mathcal {L}_{\mathcal {C}}(X^d,X)$ the subset of $\mathrm {Hom}_{\mathcal {C}}(X^d,X)$ consisting of all morphisms $f : X^d\to X$ such that the morphism $(\mathrm {Id}_{X^d}, f) : X^d \to X^d\times X = X^{d+1}$ is a Latin hypercube.

Proof. By the definition of Latin hypercubes, the morphism $\alpha = \tau ^{d+1}_{s}\circ \lambda : L\to$ $X^d$ is an isomorphism. Then setting $\iota = \lambda \circ \alpha ^{-1}$ implies immediately that $\alpha$ determines an isomorphism between the Latin hypercubes $\lambda : L\to X^{d+1}$ and $\iota : X^d\to X^{d+1}$ . We need to show that $\alpha$ and $\iota$ are unique subject to these properties. Let $\iota ' : X^d\to X^{d+1}$ a Latin hypercube and $\alpha ' : L\to X^d$ an isomorphism such that $\tau ^{d+1}_{s}\circ \iota '=$ $\mathrm {Id}_{X^d}$ and such that $\iota '\circ \alpha '= \lambda$ . Composing this equality with $\tau ^{d+1}_{s}$ yields

\begin{equation*}\alpha ' = \mathrm {Id}_{X^d}\circ \alpha ' = \tau ^{d+1}_{s}\circ \iota '\circ \alpha ' = \tau ^{d+1}_{s}\circ \lambda = \alpha \end{equation*}

This implies $\iota '=\lambda \circ \alpha ^{-1}=\iota$ , whence the uniqueness of $\iota$ and $\alpha$ as stated. The result follows.

Proof. The proof amounts to rewriting the proof of Theorem1.2, including the statement and proof of Lemma 2.1, in such a way that all steps remain valid for the Cartesian products in $\mathcal {C}$ . In order to keep this readable, we mention in each step what this corresponds to in the case where $X$ is a non-empty set (and by considering coordinates, one easily translates this to statements in $\mathcal {C}$ ).

Let $f\in$ $\mathcal {L}_{\mathcal {C}}(X^d,X)$ and $g\in$ $\mathcal {L}_{\mathcal {C}}(X^e,X)$ , where $d$ , $e$ are positive integers. That is, the morphisms

\begin{equation*}(\mathrm {Id}_{X^d}, f) \,:\, X^d \to X^{d+1}, \end{equation*}

\begin{equation*}(\mathrm {Id}_{X^e},g) \,:\, X^e \to X^{e+1}\end{equation*}

are Latin hypercubes. As in the proof of Theorem1.2, the unitality and symmetric group actions are obvious, and we only need to show, analogously to Lemma 2.1, that $(\mathrm {Id}_{X^{d+e-1}}, f\circ _i g)$ is a Latin hypercube. That is, we need to show that for $1\leq s\leq d+e-1$ and $1\leq i\leq d$ , the composition $\tau ^{d+e}_s \circ (\mathrm {Id}_{X^{d+e-1}}, f\circ _i g)$ is an automorphism of $X^{d+e-1}$ . As pointed out in Remark 2.2, since we may permute coordinates, it suffices to do this for $i=d$ and in the two cases where either $1=s\leq d-1$ or $s=d$ .

We consider first the case $1=s\leq d-1$ , so $d\geq 2$ . We need to show that the morphism

\begin{equation*}\tau ^{d+e}_1\circ (\mathrm {Id}_{X^{d+e-1}}, f\circ _d g)\end{equation*}

is an automorphism of $X^{d+e-1}$ . If $X$ is a set, then this automorphism is given by the assignment

\begin{equation*}(x_1,x_2,..,x_{d+e-1}) \mapsto (x_2,..,x_{d+e-1}, f(x_1,..,x_{d-1},g(x_d,..,x_{d+e-1}))).\end{equation*}

First, the morphism $\tau ^d_1\circ (\mathrm {Id}_{X^d}, f)$ is an automorphism of $X^d$ because $(\mathrm {Id}_{X^d},f)$ is a Latin hypercube. We note that if $X$ is a non-empty set, then the morphism $\tau ^d_1\circ (\mathrm {Id}_{X^d}, f)$ is given by the assignment

\begin{equation*}(x_1,..,x_d) \mapsto (x_2,..,x_d, f(x_1,..,x_d)).\end{equation*}

Compose this with the automorphism $\hat \sigma$ induced by the cyclic permutation $\sigma =$ $(1,2,..,d)$ on coordinates. The resulting automorphism

\begin{equation*}\hat \sigma \circ \tau ^d_1\circ (\mathrm {Id}_{X^d},f)\end{equation*}

is, for $X$ a set, given by the assignment

\begin{equation*}(x_1,..,x_d) \mapsto (f(x_1,..,x_d),x_2,..,x_d)\end{equation*}

The Cartesian product of this automorphism with $-\times \mathrm {Id}_{X^{e-1}}$ yields an automorphism of $X^{d+e-1}$ , which for $X$ a set corresponds to

\begin{equation*}(x_1,..,x_{d+e-1}) \mapsto (f(x_1,..,x_d),x_2,..,x_{d+e-1}))\end{equation*}

Again permuting cyclically, the coordinates yields an automorphism of $X^{d+e-1}$ which we will denote by $\alpha$ , which, if $X$ is a set, corresponds to

\begin{equation*}(x_1,..,x_{d+e-1}) \mapsto (x_2,..,x_{d+e-1}, f(x_1,..,x_d)).\end{equation*}

Using the fact that $\tau ^e_1\circ (\mathrm {Id}_{X^e}, g)$ is an automorphism of $X^e$ , combined with cyclically permuting coordinates, we obtain an automorphism $\gamma$ of $X^e$ , which corresponds to the assignment

\begin{equation*}(x_d,..,x_{d+e-1}) \mapsto (g(x_d,..,x_{d+e-1}), x_{d+1},..,x_{d+e-1}).\end{equation*}

Then $\beta = \mathrm {Id}_{X^{d-1}}\times \eta$ is the automorphism of $X^{d+e-1}$ which corresponds to

\begin{equation*}(x_1,..,x_{d+e-1}) \mapsto (x_1,..,x_{d-1}, g(x_d,..,x_{d+e-1}), x_{d+1},..,x_{d+e-1}).\end{equation*}

Similarly, $\gamma =\mathrm {Id}_{X^{d-1}}\times \eta \times \mathrm {Id}_X$ is an automorphism of $X^{d+e-1}$ . Now $\alpha \circ \beta$ is an automorphism of $X^{d+e-1}$ corresponding to

\begin{equation*}(x_1,..,x_{d+e-1}) \mapsto (x_2,..,x_{d-1},g(x_d,..,x_{d+e-1}),x_{d+1},..,x_{d+e-1}, f(x_1,..,x_{d-1}, g(x_d,..,x_{d+e-1}))).\end{equation*}

Thus the automorphism $\gamma ^{-1}\circ \alpha \circ \beta$ of $X^{d+e-1}$ coincides with the morphism $\tau ^{d+e}_1\circ (\mathrm {Id}_{X^{d+e-1}}, f\circ _d g)$ . This proves the result in the case $1=s\leq d-1$ .

Consider next the case $s=d$ . We need to show that the morphism $\tau ^{d+e}_d\circ (\mathrm {Id}_{X^{d+e-1}}, f\circ _d g)$ is an automorphism of $X^{d+e-1}$ . If $X$ is a set, then this automorphism is given by the assignment

\begin{equation*}(x_1,..,x_{d+e-1}) \mapsto (x_1,..,x_{d-1},x_{d+1},..,x_{d+e-1}, f(x_1,..,x_{d-1},g(x_d,..,x_{d+e-1}))).\end{equation*}

As before, by using the automorphism $\tau ^e_1\circ (\mathrm {Id}_{X^e}, g)$ , cyclically permuting the coordinates and then applying $\mathrm {Id}_{X^{d-1}}\times -$ we obtain anautomorphism $\delta$ of $X^{d+e-1}$ , which for $X$ a set corresponds to

\begin{equation*}(x_1,..,x_{d+e-1}) \mapsto (x_1,..,x_{d-1}, g(x_d,..,x_{d+e-1}), x_{d+1},..,x_{d+e-1}).\end{equation*}

Similarly, applying $-\times \mathrm {Id}_{X^{e-1}}$ to the automorphism $\tau ^d_d\circ (\mathrm {Id}_{X^d}, f)$ yields an automorphism $\epsilon$ of $X^{d+e-1}$ , which for $X$ a set corresponds to

\begin{equation*}(x_1,..,x_{d+e-1}) \mapsto (x_1,..,x_{d-1}, f(x_1,..,x_d), x_{d+1},..,x_{d+e-1}).\end{equation*}

Thus $\epsilon \circ \delta$ is the automorphism of $X^{d+e-1}$ which for $X$ a set corresponds to

\begin{equation*}(x_1,\ldots, x_{d+e-1}) \mapsto (x_1,..,x_{d-1},x_{d+1},..,x_{d+e-1}, (f\circ _d g)(x_1,..,x_{d+e-1})),\end{equation*}

and this is indeed the automorphism $\tau ^{d+e-1}_d\circ (\mathrm {Id}_{X^{d+e-1}}, f\circ _d g)$ . This proves the second case, and the result follows.

Proof. Let $\lambda \in$ $\mathrm {Hom}_{\mathcal {C}}(X^d,X)$ . By definition, we have $\lambda \in \mathcal {L}_{\mathcal {C}}(X^d, X)$ if and only if $\gamma _i\in$ $\mathrm {Aut}_{\mathcal {C}}(X^d)$ for $1\leq i\leq d$ . This is clearly equivalent to the assertion that the diagram in the statement is a pullback diagram.

Proof of Theorem 4.2 . As described in the diagram 4.1, appplied with $X^d$ instead of $X$ , there is a morphism $\mathbf {Aut}(X^d) \to \mathbf {Hom}(X^d,X^d)$ which lifts the inclusion $\mathrm {Aut}_{\mathcal {C}}(X^d)\to$ $\mathrm {Hom}_{\mathcal {C}}(X^d,X^d)$ . Both maps $\gamma$ and $\iota$ in the diagram from Lemma 4.3 lift to morphisms $\hat \gamma$ and $\hat \iota$ between the relevant internal Hom objects, and hence, by the assumptions on $\mathcal {C}$ , there is a pullback diagram in $\mathcal {C}$ of the form

4.4.

The functor $\mathrm {Hom}_{\mathcal {C}}(\unicode {x1D7D9},-)$ from $\mathcal {C}$ to the category of sets preserves pullbacks, hence sends this pullback diagram to a diagram isomorphic to that in Lemma 4.3. It follows in particular that $\mathrm {Hom}_{\mathcal {C}}(\unicode {x1D7D9}, \mathbf {L}(X^d,X))\cong$ $\mathcal {L}_{\mathcal {C}}(X^d,X)$ . The uniqueness statement follows from the fact that pullbacks are unique up to unique isomorphism. If the functor $\mathrm {Hom}_{\mathcal {C}}(\unicode {x1D7D9},-)$ is faithful, then this functor reflects monomorphisms, whence the last statement follows.

Proof. Note that since we assume $\mathrm {Hom}_{\mathcal {C}}(\unicode {x1D7D9},-)$ to be faithful (hence reflecting monomorphisms), it follows that the morphism $\mathbf {L}(X^d,X)\to$ $\mathbf {Hom}(X^d,X)$ from Theorem4.2 is a monomorphism, for any positive integer $d$ . Furthermore, as in the proofs of Theorems1.2, 3.4, showing the unitality and compatibility with symmetric group actions is straightforward. What remains to be proved is that the maps $- \circ _i -$ of the endomorphism operad induce maps on the subobjects $\mathbf {L}(X^d,X)$ of the internal Hom objects $\mathbf {Hom}(X^d,X)$ . Let $d$ , $e$ , $i$ be positive integers such that $1\leq i\leq d$ . The map

\begin{equation*} - \circ _i - : \mathrm {Hom}_{\mathcal {C}}(X^d,X) \times \mathrm {Hom}_{\mathcal {C}}(X^e,X) \to \mathrm {Hom}_{\mathcal {C}}(X^{d+e-1},X)\end{equation*}

sends $(f,g)$ to $f\circ (\mathrm {Id}_{X^{i-1}} \times g \times \mathrm {Id}_{X^{d-i}})$ . Since this involves composition and products only, this map lifts to a map of internal Hom objects

\begin{equation*} \mathbf {Hom}(X^d,X) \times \mathbf {Hom}(X^e,X) \to \mathbf {Hom}(X^{d+e-1}, X).\end{equation*}

At the level of morphism sets it follows from Theorem3.4 that we have a commutative diagram 4.6.

where the vertical maps are inclusions. We need to show that this diagram lifts to the internal Hom objects and relevant subobjects. We note that the vertical maps in the diagram 4.6 lift by Theorem 4.2, and the top horizontal map lifts by the discussion preceding the diagram 4.6. What we need to show is that the bottom horizontal map in diagram 4.6 lifts as well.

Since $\mathbf {L}(X^{d+e-1},X)$ is defined via a pullback diagram 4.4 (with $d+e-1$ instead of $d$ ), we need to show that there is a commutative diagram of the form

4.7.

Combining Lemma 4.3 (with $d+e-1$ instead of $d$ ) and diagram 4.6 yields a commutative diagram

The top horizontal and two vertical maps in this diagram lift to maps as in the diagram 4.7. The hypothesis on lifting maps through morphisms of the form $\mathbf {Aut}(Y)\to$ $\mathbf {Hom}(Y,Y)$ applied to the $d+e-1$ components on the right side of the diagram 4.7 shows the existence of the lower horizontal map making the diagram 4.7 commutative. The uniqueness of such a map follows from the fact that the right vertical map is a monomorphism, where we use that the functor $\mathrm {Hom}_{\mathcal {C}}(\unicode {x1D7D9},-)$ is faithful.

Remark 4.8. We do not know whether the hypothesis on lifting morphisms $Z\to \mathbf {Hom}(Y,Y)$ through $\mathbf {Aut}(Y)\to$ $\mathbf {Hom}(Y,Y)$ is indeed needed for Theorem 4.5 to hold. This hypothesis holds in the category of compactly generated topological spaces. It is easy to see that this hypothesis holds if $\mathbf {Aut}(Y) \to$ $\mathbf {Hom}(Y,Y)$ is a regular monomorphism (these are monomorphisms which are an equaliser of a pair of parallel morphisms), assuming as before that $\mathrm {Hom}_{\mathcal {C}}(\unicode {x1D7D9},-)$ is faithful.

Proof. Note that $L\circ _i M$ is indeed a Latin hypercube by Theorem1.2. We have

\begin{equation*}(x_1,x_2,..,x_{d+e})\in L\circ _i M\end{equation*}

if and only if

\begin{equation*}x_{d+e} = (f\circ _i g)(x_1,x_2,..,x_{d+e-1}) = f(x_1,..,x_{i-1}, g(x_i,..,x_{i+e-1}),x_{i+e},..,x_{d+e-1}).\end{equation*}

Since $z=g(x_i,..,x_{e+i-1})$ is the unique element in $X$ such that $(x_i,..,x_{i+e-1},z)\in M$ and the unique element such that $(x_1,..,x_{i-1},z,x_{i+e},..,x_{d+e})\in L$ , the first statement follows. The second statement follows from the fact that $z$ is uniquely determined by the coordinates $x_i$ , $1\leq i \leq d+e$

Proof. By permuting the coordinates it suffices to show this for $s=d$ . Let $Z$ be an object in $\mathcal {C}$ , and let $u : Z\to X^{d+1}$ and $v : Z\to X^{d-1}$ be morphisms such that

\begin{equation*}\tau ^{d+1}_d\circ u= (\mathrm {Id}_{X^{d-1}}\circ c)\circ v.\end{equation*}

We need to show that there is a unique morphism $w : Z\to X^d$ satisfying $u=(\mathrm {Id}_{X^d}\times c)\circ w$ and $v = \tau ^d_d\circ w$ . By checking on coordinates one sees that $w = (v,u_d) : Z\to X^{d-1}\times X=X^d$ is the unique morphism with this property, where as before $u_d=$ $\pi ^{d+1}_d\circ u$ .

Proof. The morphism $\tau ^{d+1}_{d+1}\circ \lambda$ is an isomorphism as part of the definition of a Latin hypercube. Let $c : \unicode {x1D7D9} \to X$ be a morphism, and let $s$ be an integer such that $1\leq s\leq d$ . Consider a pullback diagram as in the statement. We need to show that $\tau ^d_s\circ \lambda _c : L_c \to X^{d-1}$ is an isomorphism. Since $s\leq d$ and $d\geq 2$ we can complete the diagram in the statement to a commutative diagram

The right square is a pullback diagram by Lemma 5.2. Since the left square is a pullback diagram, the pasting law for pullbacks implies that the outer rectangle is a pullback diagram as well. Since the lower horizontal morphism in the outer rectangle is an isomorphism, so is the upper horizontal morphism (we use here that pullbacks preserve isomorphisms).

Remark 5.4. The notion of transversals can be adapted to the situation of Definition 1.1 as follows. Given an object $X$ in a category with finite products and a positive integer $d$ , a transversal in $X^{d+1}$ is a morphism $\sigma : X\to X^{d+1}$ such that $\sigma _i=\pi ^{d+1}_i\circ \sigma$ is an automorphism of $X$ , for $1\leq i\leq d+1$ . The morphism $\sigma$ is obviously a monomorphism. We say that such a transversal is contained in a Latin hypercube $\lambda : L\to X^{d+1}$ if there is a morphism $\iota : X\to L$ such that $\lambda \circ \iota =$ $\sigma$ . In that case, $\iota$ is necessarily a monomorphism as well. If $X$ is a non-empty set, then a transversal $\sigma$ can be identified with the subset $\sigma (X)$ of $X^{d+1}$ . If $X$ is finite and has an abelian group structure, and if $T\subseteq X^{d+1}$ is a transversal, then we have a version of the $\Delta$ -Lemma [9, Lemma2.1] as follows. Denote by $\alpha : X^{d+1}\to X$ the alternating sum map $\alpha (x_1,x_2,..,x_{d+1})=\sum _{i=1}^{d+1} (-1)^{i-1}x_i$ and by $t$ the sum of all involutions in $X$ . Since the sum of all elements in $X$ is equal to the sum $t$ of all involutions in $X$ , an elementary calculation shows that

\begin{equation*}\sum _{x \in T} \ \ \alpha (x) = \begin{cases} 0 & \text {if}\ d \ \text {is odd} \\ t & \text {if}\ d\ \text {is even.} \end{cases} \end{equation*}

If $X=\mathbb {Z}/n\mathbb {Z}$ for some positive integer $n$ , then $t=0$ if $n$ is odd, and $t=$ $\frac {n}{2} + n\mathbb {Z}$ if $n$ is even.

Remark 5.5. Adapting another well-known notion for Latin squares going back to Bose [2], the graph of a Latin hypercube $L\subseteq X^{d+1}$ of dimension $d\geq 1$ over a non-empty set $X$ is the simple graph $\Gamma (L)$ with vertex set $L$ , with an edge between two elements in $L$ if the two elements have $d-1$ coordinates at which they coincide. Being a Latin hypercube implies that if one fixes $d-1$ coordinates of an element in $L$ , then the remaining two coordinates determine each other. In particular, if two elements of $L$ coincide at $d$ coordinates, then these two elements are equal. If $|X|=n$ is finite, then $\Gamma (L)$ has $n^d$ vertices and valency $\binom {d+1}{2} (n-1)$ ; indeed, the neighbours of $(x_1,x_2,..,x_{d+1})\in L$ are obtained by first choosing a two-element subset $\{i_1,i_2\}$ of $\{1,2,..,d+1\}$ and then replacing $x_{i_1}$ by any of the $n-1$ values different from $x_{i_1}$ and replacing $x_{i_2}$ by the unique element such that the resulting $d+1$ -tuple belongs to $L$ . Adjacent vertices have $n-2 + 2 (d-1)=$ $n+2d-4$ common neighbours; indeed, two distinct elements of the form $(x_1,x_2,x_3,..,x_{d+1})$ , $(x_1',x_2',x_3,..,x_{d+1})$ in $L$ have the $n-2$ common neighbours of the form $(y_1,y_2,x_3,..,x_{d+1})$ with $y_1\neq x_1, x'_1$ (so necessarily $y_2\neq x_2,x'_2$ ), and in addition the $2(d-1)$ common neighbours of the form $(x_1,x'_2,x_3,..,y_j,..,x_{d+1})$ , $(x'_1,x_2, x_3,..,z_j,..,x_{d+1})$ in $L$ where $3\leq j\leq d+1$ and $y_j$ , $z_j$ are the uniquely determined elements by the remaining coordinates. If $d\geq 3$ , then the length of the shortest path between non-adjacent vertices depends on the number of common coordinates.

Remark 5.6. Latin hypercubes of a fixed dimension $d$ over objects in a category $\mathcal {C}$ with finite products form themselves a category. Let $X$ , $Y$ be objects in $\mathcal {C}$ . Let $\lambda : X^d\to X^{d+1}$ and $\mu : Y^d \to Y^{d+1}$ be Latin hypercubes. A morphism from $\mu$ to $\lambda$ is a morphism $\iota : Y\to X$ such that $\lambda \circ \iota ^{\times d} =$ $\iota ^{\times (d+1)}\circ \mu$ . Here $\iota ^{\times d} : Y^d\to X^d$ is the morphism obtained by taking $d$ times the product of $\iota$ ; similarly for $d+1$ . Note though that the notion of isomorphism in this category is different from the notion of isomorphism considered in the Introduction, where we regard Latin hypercubes of dimension $d$ over $X$ as objects in the under-category of $X^{d+1}$ . For a Latin hypercube $\lambda : X^d\to X^{d+1}$ of the form $\lambda = (\mathrm {Id}_{X^d}, f)$ for some morphism $f : X^d\to X$ , one checks easily that an automorphism of this Latin hypercube in the category defined here is an automorphism $\iota \in \mathrm {Aut}_{\mathcal {C}}(X)$ satisfying $\iota \circ f=$ $f\circ \iota ^{\times d}$ .