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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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