The Stacks project

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

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

Thus if $\mathcal{C}$ has countable limits then this limit and hence an object representing the functor of the lemma exist. $\square$

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

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

is commutative. Turning this diagram 90 degrees clockwise we get the desired diagram as in (2) of Categories, Definition 4.17.3. $\square$

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

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

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

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

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

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

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. $\square$

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

Here we have used the fact that

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 $\mathop{\mathrm{Hom}}\nolimits (V \amalg _ U W, T)_ n$. $\square$