## 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 $\mathop{\mathrm{Hom}}\nolimits (U, V)$ any simplicial object of $\mathcal{C}$ such that

$\mathop{\mathrm{Mor}}\nolimits _{\text{Simp}(\mathcal{C})}(W, \mathop{\mathrm{Hom}}\nolimits (U, V)) = \mathop{\mathrm{Mor}}\nolimits _{\text{Simp}(\mathcal{C})}(W \times U, V)$

functorially in the simplicial object $W$ of $\mathcal{C}$.

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

$\mathop{\mathrm{lim}}\nolimits _{\mathcal{U}^{opp}} \mathcal{V}$

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{\mathrm{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 $\mathop{\mathrm{Hom}}\nolimits (U, V)$ exists, moreover we have the expected equalities

$\mathop{\mathrm{Hom}}\nolimits (U, V)_ n = \mathop{\mathrm{Hom}}\nolimits (U \times \Delta [n], V)_0.$

Proof. We construct this simplicial object as follows. For $n \geq 0$ let $\mathop{\mathrm{Hom}}\nolimits (U, V)_ n$ denote the object of $\mathcal{C}$ representing the functor

$X \longmapsto \mathop{\mathrm{Mor}}\nolimits _{\text{Simp}(\mathcal{C})}(X \times U \times \Delta [n], V)$

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

$\mathop{\mathrm{Hom}}\nolimits (U, V)(\varphi ) : \mathop{\mathrm{Hom}}\nolimits (U, V)_ n \longrightarrow \mathop{\mathrm{Hom}}\nolimits (U, V)_ m.$

Clearly this defines a contravariant functor $\mathop{\mathrm{Hom}}\nolimits (U, V)$ from $\Delta$ into the category $\mathcal{C}$. In other words, we have a simplicial object of $\mathcal{C}$.

We have to show that $\mathop{\mathrm{Hom}}\nolimits (U, V)$ satisfies the desired universal property

$\mathop{\mathrm{Mor}}\nolimits _{\text{Simp}(\mathcal{C})}(W, \mathop{\mathrm{Hom}}\nolimits (U, V)) = \mathop{\mathrm{Mor}}\nolimits _{\text{Simp}(\mathcal{C})}(W \times U, V)$

To see this, let $f : W \to \mathop{\mathrm{Hom}}\nolimits (U, V)$ be given. We want to construct the element $f' : W \times U \to V$ of the right hand side. By construction, each $f_ n : W_ n \to \mathop{\mathrm{Hom}}\nolimits (U, V)_ n$ 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 $f : W \to \mathop{\mathrm{Hom}}\nolimits (U, V)$ 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 $f_ n : W_ n \to \mathop{\mathrm{Hom}}\nolimits (U, V)_ n$. 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

$\mathop{\mathrm{Hom}}\nolimits (V, T) \times _{\mathop{\mathrm{Hom}}\nolimits (U, T)} \mathop{\mathrm{Hom}}\nolimits (W, T) = \mathop{\mathrm{Hom}}\nolimits (V \amalg _ U W, T)$

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 $\mathop{\mathrm{Hom}}\nolimits$ 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{\mathrm{Mor}}\nolimits (X \times \Delta [n], \mathop{\mathrm{Hom}}\nolimits (V, T) \times _{\mathop{\mathrm{Hom}}\nolimits (U, T)} \mathop{\mathrm{Hom}}\nolimits (W, T)) \\ & = \mathop{\mathrm{Mor}}\nolimits (X \times \Delta [n], \mathop{\mathrm{Hom}}\nolimits (V, T)) \times _{\mathop{\mathrm{Mor}}\nolimits (X \times \Delta [n], \mathop{\mathrm{Hom}}\nolimits (U, T))} \mathop{\mathrm{Mor}}\nolimits (X \times \Delta [n], \mathop{\mathrm{Hom}}\nolimits (W, T)) \\ & = \mathop{\mathrm{Mor}}\nolimits (X \times \Delta [n] \times V, T) \times _{\mathop{\mathrm{Mor}}\nolimits (X \times \Delta [n] \times U, T)} \mathop{\mathrm{Mor}}\nolimits (X \times \Delta [n] \times W, T) \\ & = \mathop{\mathrm{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 $\mathop{\mathrm{Hom}}\nolimits (V \amalg _ U W, T)_ n$. $\square$

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