# The Stacks Project

## Tag 017H

### 14.17. Hom from simplicial sets into simplicial objects

Motivated by the discussion on internal hom we define what should be the simplicial object classifying morphisms from a simplicial set into a given simplicial object of the category $\mathcal{C}$.

Definition 14.17.1. Let $\mathcal{C}$ be a category such that the coproduct of any two objects exists. Let $U$ be a simplicial set, with $U_n$ finite nonempty for all $n \geq 0$. Let $V$ be a simplicial object of $\mathcal{C}$. We denote $\mathop{\rm Hom}\nolimits(U, V)$ any simplicial object of $\mathcal{C}$ such that $$\mathop{\rm Mor}\nolimits_{\text{Simp}(\mathcal{C})}(W, \mathop{\rm Hom}\nolimits(U, V)) = \mathop{\rm Mor}\nolimits_{\text{Simp}(\mathcal{C})}(W \times U, V)$$ functorially in the simplicial object $W$ of $\mathcal{C}$.

Of course $\mathop{\rm Hom}\nolimits(U, V)$ need not exist. Also, by the discussion in Section 14.16 we expect that if it does exist, then $\mathop{\rm Hom}\nolimits(U, V)_n = \mathop{\rm Hom}\nolimits(U \times \Delta[n], V)_0$. We do not use the italic notation for these Hom objects since $\mathop{\rm Hom}\nolimits(U, V)$ is not an internal hom.

Lemma 14.17.2. Assume the category $\mathcal{C}$ has coproducts of any two objects and countable limits. Let $U$ be a simplicial set, with $U_n$ finite nonempty for all $n \geq 0$. Let $V$ be a simplicial object of $\mathcal{C}$. Then the functor \begin{eqnarray*} \mathcal{C}^{opp} & \longrightarrow & \textit{Sets} \\ X & \longmapsto & \mathop{\rm Mor}\nolimits_{\text{Simp}(\mathcal{C})}(X \times U, V) \end{eqnarray*} is representable.

Proof. A morphism from $X \times U$ into $V$ is given by a collection of morphisms $f_u : X \to V_n$ with $n \geq 0$ and $u \in U_n$. And such a collection actually defines a morphism if and only if for all $\varphi : [m] \to [n]$ all the diagrams $$\xymatrix{ X \ar[r]^{f_u} \ar[d]_{\text{id}_X} & V_n \ar[d]^{V(\varphi)} \\ X \ar[r]^{f_{U(\varphi)(u)}} & V_m }$$ commute. Thus it is natural to introduce a category $\mathcal{U}$ and a functor $\mathcal{V} : \mathcal{U}^{opp} \to \mathcal{C}$ as follows:

1. The set of objects of $\mathcal{U}$ is $\coprod_{n \geq 0} U_n$,
2. a morphism from $u' \in U_m$ to $u \in U_n$ is a $\varphi : [m] \to [n]$ such that $U(\varphi)(u) = u'$
3. for $u \in U_n$ we set $\mathcal{V}(u) = V_n$, and
4. for $\varphi : [m] \to [n]$ such that $U(\varphi)(u) = u'$ we set $\mathcal{V}(\varphi) = V(\varphi) : V_n \to V_m$.

At this point it is clear that our functor is nothing but the functor defining $$\mathop{\rm lim}\nolimits_{\mathcal{U}^{opp}} \mathcal{V}$$ Thus if $\mathcal{C}$ has countable limits then this limit and hence an object representing the functor of the lemma exist. $\square$

Lemma 14.17.3. Assume the category $\mathcal{C}$ has coproducts of any two objects and finite limits. Let $U$ be a simplicial set, with $U_n$ finite nonempty for all $n \geq 0$. Assume that all $n$-simplices of $U$ are degenerate for all $n \gg 0$. Let $V$ be a simplicial object of $\mathcal{C}$. Then the functor \begin{eqnarray*} \mathcal{C}^{opp} & \longrightarrow & \textit{Sets} \\ X & \longmapsto & \mathop{\rm Mor}\nolimits_{\text{Simp}(\mathcal{C})}(X \times U, V) \end{eqnarray*} is representable.

Proof. We have to show that the category $\mathcal{U}$ described in the proof of Lemma 14.17.2 has a finite subcategory $\mathcal{U}'$ such that the limit of $\mathcal{V}$ over $\mathcal{U}'$ is the same as the limit of $\mathcal{V}$ over $\mathcal{U}$. We will use Categories, Lemma 4.17.4. For $m > 0$ let $\mathcal{U}_{\leq m}$ denote the full subcategory with objects $\coprod_{0 \leq n \leq m} U_m$. Let $m_0$ be an integer such that every $n$-simplex of the simplicial set $U$ is degenerate if $n > m_0$. For any $m \geq m_0$ large enough, the subcategory $\mathcal{U}_{\leq m}$ satisfies property (1) of Categories, Definition 4.17.3.

Suppose that $u \in U_n$ and $u' \in U_{n'}$ with $n, n' \leq m_0$ and suppose that $\varphi : [k] \to [n]$, $\varphi' : [k] \to [n']$ are morphisms such that $U(\varphi)(u) = U(\varphi')(u')$. A simple combinatorial argument shows that if $k > 2m_0$, then there exists an index $0 \leq i \leq 2m_0$ such that $\varphi(i) =\varphi(i + 1)$ and $\varphi'(i) = \varphi'(i + 1)$. (The pigeon hole principle would tell you this works if $k > m_0^2$ which is good enough for the argument below anyways.) Hence, if $k > 2m_0$, we may write $\varphi = \psi \circ \sigma^{k - 1}_i$ and $\varphi' = \psi' \circ \sigma^{k - 1}_i$ for some $\psi : [k - 1] \to [n]$ and some $\psi' : [k - 1] \to [n']$. Since $s^{k - 1}_i : U_{k - 1} \to U_k$ is injective, see Lemma 14.3.6, we conclude that $U(\psi)(u) = U(\psi')(u')$ also. Continuing in this fashion we conclude that given morphisms $u \to z$ and $u' \to z$ of $\mathcal{U}$ with $u, u' \in \mathcal{U}_{\leq m_0}$, there exists a commutative diagram $$\xymatrix{ u \ar[rd] \ar[rrd] & & \\ & a \ar[r] & z \\ u' \ar[ru] \ar[rru] }$$ with $a \in \mathcal{U}_{\leq 2m_0}$.

It is easy to deduce from this that the finite subcategory $\mathcal{U}_{\leq 2m_0}$ works. Namely, suppose given $x' \in U_n$ and $x'' \in U_{n'}$ with $n, n' \leq 2m_0$ as well as morphisms $x' \to x$ and $x'' \to x$ of $\mathcal{U}$ with the same target. By our choice of $m_0$ we can find objects $u, u'$ of $\mathcal{U}_{\leq m_0}$ and morphisms $u \to x'$, $u' \to x''$. By the above we can find $a \in \mathcal{U}_{\leq 2m_0}$ and morphisms $u \to a$, $u' \to a$ such that $$\xymatrix{ u \ar[rd] \ar[rrd] \ar[r] & x' \ar[rd] & \\ & a \ar[r] & x \\ u' \ar[ru] \ar[rru] \ar[r] & x'' \ar[ru] & }$$ is commutative. Turning this diagram 90 degrees clockwise we get the desired diagram as in (2) of Categories, Definition 4.17.3. $\square$

Lemma 14.17.4. Assume the category $\mathcal{C}$ has coproducts of any two objects and finite limits. Let $U$ be a simplicial set, with $U_n$ finite nonempty for all $n \geq 0$. Assume that all $n$-simplices of $U$ are degenerate for all $n \gg 0$. Let $V$ be a simplicial object of $\mathcal{C}$. Then $\mathop{\rm Hom}\nolimits(U, V)$ exists, moreover we have the expected equalities $$\mathop{\rm Hom}\nolimits(U, V)_n = \mathop{\rm Hom}\nolimits(U \times \Delta[n], V)_0.$$

Proof. We construct this simplicial object as follows. For $n \geq 0$ let $\mathop{\rm Hom}\nolimits(U, V)_n$ denote the object of $\mathcal{C}$ representing the functor $$X \longmapsto \mathop{\rm Mor}\nolimits_{\text{Simp}(\mathcal{C})}(X \times U \times \Delta[n], V)$$ This exists by Lemma 14.17.3 because $U \times \Delta[n]$ is a simplicial set with finite sets of simplices and no nondegenerate simplices in high enough degree, see Lemma 14.11.5. For $\varphi : [m] \to [n]$ we obtain an induced map of simplicial sets $\varphi : \Delta[m] \to \Delta[n]$. Hence we obtain a morphism $X \times U \times \Delta[m] \to X \times U \times \Delta[n]$ functorial in $X$, and hence a transformation of functors, which in turn gives $$\mathop{\rm Hom}\nolimits(U, V)(\varphi) : \mathop{\rm Hom}\nolimits(U, V)_n \longrightarrow \mathop{\rm Hom}\nolimits(U, V)_m.$$ Clearly this defines a contravariant functor $\mathop{\rm Hom}\nolimits(U, V)$ from $\Delta$ into the category $\mathcal{C}$. In other words, we have a simplicial object of $\mathcal{C}$.

We have to show that $\mathop{\rm Hom}\nolimits(U, V)$ satisfies the desired universal property $$\mathop{\rm Mor}\nolimits_{\text{Simp}(\mathcal{C})}(W, \mathop{\rm Hom}\nolimits(U, V)) = \mathop{\rm Mor}\nolimits_{\text{Simp}(\mathcal{C})}(W \times U, V)$$ To see this, let $f : W \to \mathop{\rm Hom}\nolimits(U, V)$ be given. We want to construct the element $f' : W \times U \to V$ of the right hand side. By construction, each $f_n : W_n \to \mathop{\rm Hom}\nolimits(U, V)_n$ corresponds to a morphism $f_n : W_n \times U \times \Delta[n] \to V$. Further, for every morphism $\varphi : [m] \to [n]$ the diagram $$\xymatrix{ W_n \times U \times \Delta[m] \ar[rr]_{W(\varphi)\times \text{id} \times \text{id}} \ar[d]_{\text{id} \times \text{id} \times \varphi} & & W_m \times U \times \Delta[m] \ar[d]^{f_m} \\ W_n \times U \times \Delta[n] \ar[rr]^{f_n} & & V }$$ is commutative. For $\psi : [n] \to [k]$ in $(\Delta[n])_k$ we denote $(f_n)_{k, \psi} : W_n \times U_k \to V_k$ the component of $(f_n)_k$ corresponding to the element $\psi$. We define $f'_n : W_n \times U_n \to V_n$ as $f'_n = (f_n)_{n, \text{id}}$, in other words, as the restriction of $(f_n)_n : W_n \times U_n \times (\Delta[n])_n \to V_n$ to $W_n \times U_n \times \text{id}_{[n]}$. To see that the collection $(f'_n)$ defines a morphism of simplicial objects, we have to show for any $\varphi : [m] \to [n]$ that $V(\varphi) \circ f'_n = f'_m \circ W(\varphi) \times U(\varphi)$. The commutative diagram above says that $(f_n)_{m, \varphi} : W_n \times U_m \to V_m$ is equal to $(f_m)_{m, \text{id}} \circ W(\varphi) : W_n \times U_m \to V_m$. But then the fact that $f_n$ is a morphism of simplicial objects implies that the diagram $$\xymatrix{ W_n \times U_n \times (\Delta[n])_n \ar[r]_-{(f_n)_n} \ar[d]_{\text{id} \times U(\varphi) \times \varphi} & V_n \ar[d]^{V(\varphi)} \\ W_n \times U_m \times (\Delta[n])_m \ar[r]^-{(f_n)_m} & V_m }$$ is commutative. And this implies that $(f_n)_{m, \varphi} \circ U(\varphi)$ is equal to $V(\varphi) \circ (f_n)_{n, \text{id}}$. Altogether we obtain $V(\varphi) \circ (f_n)_{n, \text{id}} = (f_n)_{m, \varphi} \circ U(\varphi) = (f_m)_{m, \text{id}} \circ W(\varphi)\circ U(\varphi) = (f_m)_{m, \text{id}} \circ W(\varphi)\times U(\varphi)$ as desired.

On the other hand, given a morphism $f' : W \times U \to V$ we define a morphism $f : W \to \mathop{\rm Hom}\nolimits(U, V)$ as follows. By Lemma 14.13.4 the morphisms $\text{id} : W_n \to W_n$ corresponds to a unique morphism $c_n : W_n \times \Delta[n] \to W$. Hence we can consider the composition $$W_n \times \Delta[n] \times U \xrightarrow{c_n} W \times U \xrightarrow{f'} V.$$ By construction this corresponds to a unique morphism $f_n : W_n \to \mathop{\rm Hom}\nolimits(U, V)_n$. We leave it to the reader to see that these define a morphism of simplicial sets as desired.

We also leave it to the reader to see that $f \mapsto f'$ and $f' \mapsto f$ are mutually inverse operations. $\square$

Lemma 14.17.5. Assume the category $\mathcal{C}$ has coproducts of any two objects and finite limits. Let $a : U \to V$, $b : U \to W$ be morphisms of simplicial sets. Assume $U_n, V_n, W_n$ finite nonempty for all $n \geq 0$. Assume that all $n$-simplices of $U, V, W$ are degenerate for all $n \gg 0$. Let $T$ be a simplicial object of $\mathcal{C}$. Then $$\mathop{\rm Hom}\nolimits(V, T) \times_{\mathop{\rm Hom}\nolimits(U, T)} \mathop{\rm Hom}\nolimits(W, T) = \mathop{\rm Hom}\nolimits(V \amalg_U W, T)$$ In other words, the fibre product on the left hand side is represented by the Hom object on the right hand side.

Proof. By Lemma 14.17.4 all the required $\mathop{\rm Hom}\nolimits$ objects exist and satisfy the correct functorial properties. Now we can identify the $n$th term on the left hand side as the object representing the functor that associates to $X$ the first set of the following sequence of functorial equalities \begin{align*} & \mathop{\rm Mor}\nolimits(X \times \Delta[n], \mathop{\rm Hom}\nolimits(V, T) \times_{\mathop{\rm Hom}\nolimits(U, T)} \mathop{\rm Hom}\nolimits(W, T)) \\ & = \mathop{\rm Mor}\nolimits(X \times \Delta[n], \mathop{\rm Hom}\nolimits(V, T)) \times_{\mathop{\rm Mor}\nolimits(X \times \Delta[n], \mathop{\rm Hom}\nolimits(U, T))} \mathop{\rm Mor}\nolimits(X \times \Delta[n], \mathop{\rm Hom}\nolimits(W, T)) \\ & = \mathop{\rm Mor}\nolimits(X \times \Delta[n] \times V, T) \times_{\mathop{\rm Mor}\nolimits(X \times \Delta[n] \times U, T)} \mathop{\rm Mor}\nolimits(X \times \Delta[n] \times W, T) \\ & = \mathop{\rm Mor}\nolimits(X \times \Delta[n] \times (V \amalg_U W), T)) \end{align*} Here we have used the fact that $$(X \times \Delta[n] \times V) \times_{X \times \Delta[n] \times U} (X \times \Delta[n] \times W) = X \times \Delta[n] \times (V \amalg_U W)$$ which is easy to verify term by term. The result of the lemma follows as the last term in the displayed sequence of equalities corresponds to $\mathop{\rm Hom}\nolimits(V \amalg_U W, T)_n$. $\square$

The code snippet corresponding to this tag is a part of the file simplicial.tex and is located in lines 1364–1741 (see updates for more information).

\section{Hom from simplicial sets into simplicial objects}
\label{section-hom-from-simplicial-sets}

\noindent
Motivated by the discussion on internal hom we define
what should be the simplicial object classifying
morphisms from a simplicial set into a given
simplicial object of the category $\mathcal{C}$.

\begin{definition}
\label{definition-hom-from-simplicial-set}
Let $\mathcal{C}$ be a category such that the coproduct
of any two objects exists.
Let $U$ be a simplicial set, with $U_n$ finite nonempty
for all $n \geq 0$.
Let $V$ be a simplicial object of $\mathcal{C}$.
We denote {\it $\Hom(U, V)$} any simplicial object of
$\mathcal{C}$ such that
$$\Mor_{\text{Simp}(\mathcal{C})}(W, \Hom(U, V)) = \Mor_{\text{Simp}(\mathcal{C})}(W \times U, V)$$
functorially in the simplicial object $W$ of $\mathcal{C}$.
\end{definition}

\noindent
Of course $\Hom(U, V)$ need not exist.
Also, by the discussion in Section \ref{section-internal-hom}
we expect that if it does exist, then
$\Hom(U, V)_n = \Hom(U \times \Delta[n], V)_0$.
We do not use the italic notation for these Hom objects
since $\Hom(U, V)$ is not an internal hom.

\begin{lemma}
\label{lemma-exists-hom-0-from-simplicial-set}
Assume the category $\mathcal{C}$
has coproducts of any two objects and countable
limits. Let $U$ be a simplicial set, with $U_n$ finite nonempty
for all $n \geq 0$.
Let $V$ be a simplicial object of $\mathcal{C}$.
Then the functor
\begin{eqnarray*}
\mathcal{C}^{opp} & \longrightarrow & \textit{Sets} \\
X
& \longmapsto &
\Mor_{\text{Simp}(\mathcal{C})}(X \times U, V)
\end{eqnarray*}
is representable.
\end{lemma}

\begin{proof}
A morphism from $X \times U$ into $V$ is given by a collection
of morphisms $f_u : X \to V_n$ with $n \geq 0$ and $u \in U_n$.
And such a collection actually defines a morphism if and only
if for all $\varphi : [m] \to [n]$ all the diagrams
$$\xymatrix{ X \ar[r]^{f_u} \ar[d]_{\text{id}_X} & V_n \ar[d]^{V(\varphi)} \\ X \ar[r]^{f_{U(\varphi)(u)}} & V_m }$$
commute. Thus it is natural to introduce a category
$\mathcal{U}$ and a functor
$\mathcal{V} : \mathcal{U}^{opp} \to \mathcal{C}$
as follows:
\begin{enumerate}
\item The set of objects of $\mathcal{U}$ is
$\coprod_{n \geq 0} U_n$,
\item a morphism from $u' \in U_m$ to $u \in U_n$
is a $\varphi : [m] \to [n]$ such that $U(\varphi)(u) = u'$
\item for $u \in U_n$ we set $\mathcal{V}(u) = V_n$, and
\item for $\varphi : [m] \to [n]$ such that $U(\varphi)(u) = u'$
we set $\mathcal{V}(\varphi) = V(\varphi) : V_n \to V_m$.
\end{enumerate}
At this point it is clear that our functor is nothing but the
functor defining
$$\lim_{\mathcal{U}^{opp}} \mathcal{V}$$
Thus if $\mathcal{C}$ has countable limits then this limit
and hence an object representing the functor of the lemma
exist.
\end{proof}

\begin{lemma}
\label{lemma-exists-hom-0-from-simplicial-set-finite}
Assume the category $\mathcal{C}$
has coproducts of any two objects and finite
limits. Let $U$ be a simplicial set, with $U_n$ finite nonempty
for all $n \geq 0$. Assume that all $n$-simplices
of $U$ are degenerate for all $n \gg 0$.
Let $V$ be a simplicial object of $\mathcal{C}$.
Then the functor
\begin{eqnarray*}
\mathcal{C}^{opp} & \longrightarrow & \textit{Sets} \\
X
& \longmapsto &
\Mor_{\text{Simp}(\mathcal{C})}(X \times U, V)
\end{eqnarray*}
is representable.
\end{lemma}

\begin{proof}
We have to show that the category $\mathcal{U}$ described
in the proof of Lemma \ref{lemma-exists-hom-0-from-simplicial-set}
has a finite subcategory $\mathcal{U}'$ such that the limit
of $\mathcal{V}$ over $\mathcal{U}'$ is the same as the
limit of $\mathcal{V}$ over $\mathcal{U}$. We will use
Categories, Lemma \ref{categories-lemma-initial}.
For $m > 0$ let $\mathcal{U}_{\leq m}$ denote the full
subcategory with objects $\coprod_{0 \leq n \leq m} U_m$.
Let $m_0$ be an integer such that every $n$-simplex
of the simplicial set $U$ is degenerate if $n > m_0$.
For any $m \geq m_0$ large enough, the subcategory
$\mathcal{U}_{\leq m}$ satisfies property (1) of
Categories, Definition \ref{categories-definition-initial}.

\medskip\noindent
Suppose that $u \in U_n$ and
$u' \in U_{n'}$ with $n, n' \leq m_0$ and suppose that
$\varphi : [k] \to [n]$, $\varphi' : [k] \to [n']$
are morphisms such that $U(\varphi)(u) = U(\varphi')(u')$.
A simple combinatorial argument shows that if $k > 2m_0$,
then there exists an index $0 \leq i \leq 2m_0$ such that
$\varphi(i) =\varphi(i + 1)$ and $\varphi'(i) = \varphi'(i + 1)$.
(The pigeon hole principle would tell you this works if
$k > m_0^2$ which is good enough for the argument below
anyways.) Hence, if $k > 2m_0$, we may write
$\varphi = \psi \circ \sigma^{k - 1}_i$ and
$\varphi' = \psi' \circ \sigma^{k - 1}_i$ for some
$\psi : [k - 1] \to [n]$ and some $\psi' : [k - 1] \to [n']$.
Since $s^{k - 1}_i : U_{k - 1} \to U_k$ is injective,
see Lemma \ref{lemma-si-injective}, we conclude that
$U(\psi)(u) = U(\psi')(u')$ also. Continuing in this
fashion we conclude that given morphisms
$u \to z$ and $u' \to z$ of $\mathcal{U}$
with $u, u' \in \mathcal{U}_{\leq m_0}$, there exists
a commutative diagram
$$\xymatrix{ u \ar[rd] \ar[rrd] & & \\ & a \ar[r] & z \\ u' \ar[ru] \ar[rru] }$$
with $a \in \mathcal{U}_{\leq 2m_0}$.

\medskip\noindent
It is easy to deduce from this that the finite subcategory
$\mathcal{U}_{\leq 2m_0}$ works. Namely, suppose given
$x' \in U_n$ and $x'' \in U_{n'}$ with $n, n' \leq 2m_0$ as well as
morphisms $x' \to x$ and $x'' \to x$ of $\mathcal{U}$
with the same target. By our choice of $m_0$ we can
find objects $u, u'$ of $\mathcal{U}_{\leq m_0}$ and
morphisms $u \to x'$, $u' \to x''$.
By the above we can find $a \in \mathcal{U}_{\leq 2m_0}$
and morphisms $u \to a$, $u' \to a$ such that
$$\xymatrix{ u \ar[rd] \ar[rrd] \ar[r] & x' \ar[rd] & \\ & a \ar[r] & x \\ u' \ar[ru] \ar[rru] \ar[r] & x'' \ar[ru] & }$$
is commutative. Turning this diagram 90 degrees clockwise
we get the desired diagram as in (2) of
Categories, Definition \ref{categories-definition-initial}.
\end{proof}

\begin{lemma}
\label{lemma-exists-hom-from-simplicial-set-finite}
Assume the category $\mathcal{C}$
has coproducts of any two objects and finite
limits. Let $U$ be a simplicial set, with $U_n$ finite nonempty
for all $n \geq 0$. Assume that all $n$-simplices
of $U$ are degenerate for all $n \gg 0$.
Let $V$ be a simplicial object of $\mathcal{C}$.
Then $\Hom(U, V)$ exists, moreover
we have the expected equalities
$$\Hom(U, V)_n = \Hom(U \times \Delta[n], V)_0.$$
\end{lemma}

\begin{proof}
We construct this simplicial object as follows.
For $n \geq 0$ let $\Hom(U, V)_n$ denote
the object of $\mathcal{C}$ representing the
functor
$$X \longmapsto \Mor_{\text{Simp}(\mathcal{C})}(X \times U \times \Delta[n], V)$$
This exists by Lemma \ref{lemma-exists-hom-0-from-simplicial-set-finite}
because $U \times \Delta[n]$ is a simplicial set with finite
sets of simplices and no nondegenerate simplices in high enough degree,
see Lemma \ref{lemma-product-degenerate}.
For $\varphi : [m] \to [n]$ we obtain an induced map of simplicial
sets $\varphi : \Delta[m] \to \Delta[n]$. Hence we obtain a morphism
$X \times U \times \Delta[m] \to X \times U \times \Delta[n]$
functorial in $X$, and hence a transformation of functors,
which in turn gives
$$\Hom(U, V)(\varphi) : \Hom(U, V)_n \longrightarrow \Hom(U, V)_m.$$
Clearly this defines a contravariant functor
$\Hom(U, V)$ from
$\Delta$ into the category $\mathcal{C}$.
In other words, we have a simplicial object of $\mathcal{C}$.

\medskip\noindent
We have to show that $\Hom(U, V)$ satisfies the desired
universal property
$$\Mor_{\text{Simp}(\mathcal{C})}(W, \Hom(U, V)) = \Mor_{\text{Simp}(\mathcal{C})}(W \times U, V)$$
To see this, let $f : W \to \Hom(U, V)$ be given.
We want to construct the element $f' : W \times U \to V$
of the right hand side.
By construction, each $f_n : W_n \to \Hom(U, V)_n$
corresponds to a morphism
$f_n : W_n \times U \times \Delta[n] \to V$. Further,
for every morphism $\varphi : [m] \to [n]$ the
diagram
$$\xymatrix{ W_n \times U \times \Delta[m] \ar[rr]_{W(\varphi)\times \text{id} \times \text{id}} \ar[d]_{\text{id} \times \text{id} \times \varphi} & & W_m \times U \times \Delta[m] \ar[d]^{f_m} \\ W_n \times U \times \Delta[n] \ar[rr]^{f_n} & & V }$$
is commutative. For $\psi : [n] \to [k]$ in $(\Delta[n])_k$
we denote $(f_n)_{k, \psi} : W_n \times U_k \to V_k$
the component of $(f_n)_k$ corresponding to the element
$\psi$. We define $f'_n : W_n \times U_n \to V_n$
as $f'_n = (f_n)_{n, \text{id}}$, in other words, as
the restriction of
$(f_n)_n : W_n \times U_n \times (\Delta[n])_n \to V_n$
to $W_n \times U_n \times \text{id}_{[n]}$.
To see that the collection $(f'_n)$ defines a
morphism of simplicial objects, we have to show
for any $\varphi : [m] \to [n]$ that
$V(\varphi) \circ f'_n = f'_m \circ W(\varphi) \times U(\varphi)$.
The commutative diagram above says that
$(f_n)_{m, \varphi} : W_n \times U_m \to V_m$
is equal to
$(f_m)_{m, \text{id}} \circ W(\varphi) : W_n \times U_m \to V_m$.
But then the fact that $f_n$ is a morphism of simplicial
objects implies that the diagram
$$\xymatrix{ W_n \times U_n \times (\Delta[n])_n \ar[r]_-{(f_n)_n} \ar[d]_{\text{id} \times U(\varphi) \times \varphi} & V_n \ar[d]^{V(\varphi)} \\ W_n \times U_m \times (\Delta[n])_m \ar[r]^-{(f_n)_m} & V_m }$$
is commutative. And this implies that
$(f_n)_{m, \varphi} \circ U(\varphi)$ is
equal to $V(\varphi) \circ (f_n)_{n, \text{id}}$.
Altogether we obtain
$V(\varphi) \circ (f_n)_{n, \text{id}} = (f_n)_{m, \varphi} \circ U(\varphi) = (f_m)_{m, \text{id}} \circ W(\varphi)\circ U(\varphi) = (f_m)_{m, \text{id}} \circ W(\varphi)\times U(\varphi)$
as desired.

\medskip\noindent
On the other hand, given a morphism
$f' : W \times U \to V$ we define
a morphism $f : W \to \Hom(U, V)$
as follows. By Lemma \ref{lemma-morphism-from-coproduct} the morphisms
$\text{id} : W_n \to W_n$ corresponds to a unique
morphism $c_n : W_n \times \Delta[n] \to W$.
Hence we can consider the composition
$$W_n \times \Delta[n] \times U \xrightarrow{c_n} W \times U \xrightarrow{f'} V.$$
By construction this corresponds to a unique morphism
$f_n : W_n \to \Hom(U, V)_n$. We leave it to the reader
to see that these define a morphism of simplicial sets as
desired.

\medskip\noindent
We also leave it to the reader to see that
$f \mapsto f'$ and $f' \mapsto f$ are mutually inverse
operations.
\end{proof}

\begin{lemma}
\label{lemma-hom-from-coprod}
Assume the category $\mathcal{C}$
has coproducts of any two objects and finite
limits. Let $a : U \to V$, $b : U \to W$
be morphisms of simplicial sets.
Assume $U_n, V_n, W_n$ finite nonempty for all $n \geq 0$.
Assume that all $n$-simplices of $U, V, W$
are degenerate for all $n \gg 0$.
Let $T$ be a simplicial object of $\mathcal{C}$.
Then
$$\Hom(V, T) \times_{\Hom(U, T)} \Hom(W, T) = \Hom(V \amalg_U W, T)$$
In other words, the fibre product on the left hand
side is represented by the Hom object on the right hand side.
\end{lemma}

\begin{proof}
By Lemma \ref{lemma-exists-hom-from-simplicial-set-finite}
all the required $\Hom$ objects exist and satisfy the
correct functorial properties. Now we can identify
the $n$th term on the left hand side as the object
representing the functor that associates to $X$
the first set of the following sequence of functorial
equalities
\begin{align*}
&
\Mor(X \times \Delta[n],
\Hom(V, T) \times_{\Hom(U, T)} \Hom(W, T)) \\
& =
\Mor(X \times \Delta[n], \Hom(V, T))
\times_{\Mor(X \times \Delta[n], \Hom(U, T))}
\Mor(X \times \Delta[n], \Hom(W, T)) \\
& =
\Mor(X \times \Delta[n] \times V, T)
\times_{\Mor(X \times \Delta[n] \times U, T)}
\Mor(X \times \Delta[n] \times W, T) \\
& =
\Mor(X \times \Delta[n] \times (V \amalg_U W), T))
\end{align*}
Here we have used the fact that
$$(X \times \Delta[n] \times V) \times_{X \times \Delta[n] \times U} (X \times \Delta[n] \times W) = X \times \Delta[n] \times (V \amalg_U W)$$
which is easy to verify term by term. The result of the lemma
follows as the last term in the displayed sequence of
equalities corresponds to $\Hom(V \amalg_U W, T)_n$.
\end{proof}

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