Simplicial Proof of the Homotopy Colimit Theorem

The Main Theorem

The Grothendieck construction

\begin{align*} \int_J: (\mathsf{Cat}_{\text{Thom}})^J \to \mathsf{Cat}_{\text{Thom}} \end{align*}

is a model for the homotopy colimit functor in the Thomason model structure $\mathsf{Cat}_{\text{Thom}}$. In other words we are showing that given a functor $F: J \to \mathsf{Cat}$ there is a natural homotopy equivalence

\begin{align*} \eta: \text{hocolim}NF \to N\int_J F\end{align*}

of the homotopy colimit of $NF$ and the nerve of the Grothendieck construction.

Outline

  1. The Grothendieck construction
  2. Simplicial proof of the Homotopy Colimit Theorem
    • Defining a map $\eta: \text{hocolim}NF \to N\int_J F$
    • Construct a new functor $\tilde{F}: J \to \mathsf{Cat}$
    • Produce natural homotopy equivalences
      \begin{align*} \text{hocolim}NF \leftarrow_{\lambda_1} N\tilde{F} \rightarrow_{\lambda_2} N\int_J F \end{align*}
    • Construct a simplicial homotopy $H: \eta \lambda_1 \cong \lambda_2$ which implies that $\eta$ is a homotopy equivalence since $\lambda_1$ and $\lambda_2$ are.

The Grothendieck Construction

Let $F: J \to \mathsf{Cat}$ be a functor. J is a fixed small category. The Grothendieck construction on $F$, is a category with the following data:

  • objects: $(j,x)$ where $j$ is an object of $J$ and $x$ is an object of $F(J)$
  • morphisms: $(s,u): (j,x) \to (j',x')$ where $s: j \to j'$ a morphism in J and $u: F(s)(x) \to x'$
    • composition: $(s, u) \circ (s',u') = (s' \circ s, u' \circ F(s'(u)))$
    • identity at $(j,x)$ is given by $(id_j, id_x)$

This construction extends to a functor $\int_J: \mathsf{Cat}^J \to \mathsf{Cat}$. Saying just this is a bit opaque so I will explain a little bit further. Suyppose that we have a natural transformation of functors $f: F \implies F'$ where $F,F': J \to \mathsf{Cat}$. This natural transformation induces a map

\begin{align*} \int_J f: \int_J F \to \int_J F' \end{align*}

such that $\int_J f(j,x) = (j, f(j)(x))$ and $\int_J f(s,u) = (s, f(j')(u))$. Therefore, $\int_J$ is a functor from the functor category $\mathsf{Cat}^J$.

The Simplicial Proof

We first show that there is indeed a natural map $\eta: \text{hocolim}NF \to N\int_J F$. By the simplicial replacement lemma we know that $\text{hocolim}NF$ is the diagonal of a bisimplicial set $\sqcup_*NF$. We start by describing a $(p,q)$-simplex of $\sqcup_* NF$. These are:

  • a string of $p$ composible morphisms of $\mathsf{J}$: $J_0 \leftarrow_{j_1} J_1 \leftarrow_{j_2} \leftarrow \dots \leftarrow_{j_p} J_p$ with a string of $q$ composible morphisms of $F(J_p)$: $X_0 \leftarrow_{x_1} X_1 \leftarrow_{x_2} \dots \leftarrow_{x_q} X_q$.

The simplicial operators $(d_i, 1)$, $(s_i, 1)$, $(1, d_i)$, $(1, s_i)$ act as they do in $N$(by composing morphisms) except for one:

\begin{align*}(d_p, 1)(J_0 \leftarrow_{j_1} J_1 \leftarrow_{j_2} \leftarrow \dots \leftarrow_{j_p} J_p, X_0 \leftarrow_{x_1} X_1 \leftarrow_{x_2} \dots \leftarrow_{x_q} X_q) \\ = (J_0 \leftarrow_{j_1} \dots \leftarrow_{j_{p-1}} J_p, F(s_p)(X_0) \leftarrow_{F(j_p)(x_1)} \dots \leftarrow F(j_p)(X_p)) \end{align*}

We can now define $\eta$ on the $p-$simplices in $(\text{hocolim}NF)_* = \sqcup_*NF(p,p)$ as follows:

\begin{align*} &\eta_p(J_0 \leftarrow_{j_1} \dots \leftarrow_{j_p} J_p, X_0 \leftarrow _{x_1} \dots \leftarrow_{x_p}X_p) \\ &= (J_0, F(j_1\dots j_p)(X_0)) \longleftarrow_{(j_1, F(j_1\dots j_p)(x_1))} \dots \\ &\dots \leftarrow (J_1, F(j_1\dots j_p)(X_1)) \leftarrow \dots \longleftarrow_{(j_p, F(j_p)(x_p))} (K_p, F(j_p)(X_p))\end{align*}

Note that $\eta_p$ is a simplicial map that takes a $(p,p)$ simplex in $\sqcup_* NF$ and maps it to an element of $N_p(\int_J F)$. Now that we have constructed $\eta_p$ we want to construct a functor $\tilde{F}: J \to \mathsf{Cat}$ based off of the functor $F$. Let $\tilde{F}(J) = \pi / J$ where $\pi: \int_J F \to J$ such that $\pi(j, x) = j$ and $\pi(s,u) = s$. In more detail, $\tilde{F}(K)$ is a category as follows:

  • objects: $(l,x)$ where $l: L \to J$ a morphism in $\mathsf{J}$ and $x$ and object of $F(L)$
  • morphisms: $(s_1, u_1): (l,x) \to (l', x')$ where $s_1: L \to L'$ such that $l = l' \circ s_1$ and $u_1: F(s_1)(x) \to x'$
    • composition: $(s_1, u_1) \circ (s_2, u_2) = (s_1 \circ s_2, u_1 \circ F(s_2)(u_2))$

A morphism $j: J \to J'$ in $\mathsf{J}$ induces a functor $\tilde{F}(s): \tilde{F}(J) \to \tilde{F}(J')$ such that:

  • on objects: $\tilde{F}F(s)(l, x) = (sl, x)$
  • on morphisms: $\tilde{F}(s)(s_1, u_1) = (s_1, u_1)$ Therefore, $\tilde{F}: \mathsf{J} \to \mathsf{Cat}$ is a functor.

Next, we need to show that there is a natural homotopy equivalence

\begin{align*} \lambda_1: \text{hocolim}N\tilde{F} \to \text{hocolim}NF \end{align*}

To do this, note that there exists a canonical functor

\begin{align*} \tilde{F}(J) \to F(J) \quad (l,x) &\mapsto F(l)(x) \\ (s,u) &\mapsto F(l')(x) \end{align*}

where $(s,u): (l, x) \to (l', x')$. This functor has a right adjoint

\begin{align*} F(J) \to \tilde{F}(J) \quad x \mapsto (1,x) \end{align*}

If a functor has a left or right adjoint then it is a homotopy equivalence meaning that it follows that $N\tilde{F}(J) \to NF(J)$ is a homotopy equivalence. We also know that the functor $\tilde{F}(J) \to F(J)$ gives a natural transfrom $\tilde{F} \implies F$ of functors $\mathsf{J} \to \mathsf{Cat}$ which implies that $N\tilde{F} \to NF$ is a natural equivalence. This induces a natural homotopy equivalence $\lambda_1: \text{hocolim}N\tilde{F} \to \text{hocolim}(NF)$. Now that we have shown that $\lambda_1$ is a natural homotopy equivalence, we need to show that $\lambda_2: \text{hocolim}N\tilde{F} \to N(\int_J F)$ is a natural equivalence. To do this, we first recall that the nerve of a category $N\mathsf{C}$ is defined to be the simplicial set whose $n$-simplices are strings of $n-$many composable maps in $\mathsf{C}$. This means that a $p-$simplex of $N\tilde{F}(\mathsf{J})$ can be identified with a string of $p$-morphisms that live in $\int_{J}F$:

\begin{align*} (L_0, x_0) \leftarrow_{(s_1, u_1)} \leftarrow \dots \leftarrow (L_q, x_q)\end{align*}

together with a map $l: L_0 \to J$. This corresponds to the $p-$simplex given by the string of $p$morphisms in $\tilde{F}(J)$:

\begin{align*} (l,x_0) \leftarrow_{s_1, u_1} (ll_1, X_1) \leftarrow \dots \leftarrow (ll_1\dots l_q, X_q) \end{align*}.

This means that the bisimplicial set $\sqcup_* N\tilde{F}$ has the following data:

\begin{align*} J_0 \leftarrow_{j_1} \dots \leftarrow_{j_p} J_p, J_p \leftarrow_l L_0, (L_0, X_0) \leftarrow \dots \leftarrow (L_q, X_q)\end{align*}

Therefore, the map $\lambda_2: \text{diag}\sqcup_* N\tilde{F} \to N(\int_J f)$ will send a $(q,q)$-simplex to the $q-$simplex $(L_0, X_0) \leftarrow \dots \leftarrow (L_q, X_q)$ of $N(\int_J F)$. So now we have the $\lambda_2$, but how do we know that this is the correct one? Well we want to show that it's a homotopy equivalence. To do this we want to think of $N\int_J F$ as a bisimplicial set where the $(p,q)$ simplicies are constant in the $p$ direction(i.e. $(N\int_J F)_{(p,q)} = (N\int_J F)_q$). Note that with this notation we have $(N\int_JF)_{*} = \text{diag}(N\int_J F)_{**}$. Then note that since $\lambda_2$ sends a $(q,q)$ simplex to a $q$ simplex we get that $\lambda_2$ must be the diagonalization of the bisimplicial map

\begin{align*} \Lambda: \bigsqcup_* N\tilde{F} \to N(\int_J F)_{**}\end{align*}

This mainly comes from the fact that $(\text{hocolim} NF)_* = \sqcup_* NF(p,p)$ and the above fact that $(N \int_J F)_* = \text{diag} (N\int_J F)_{**}$. Now we want to recall two facts about bisimplicial sets, the proof of which can be found in here.

  1. $|p \to \text{diag}T_p| \cong |q \to |p \to T_{p,q}||$ for a bisimplicial $T_{**}$. Here $|\cdot|$ is the geometric realization.
  2. Given a bisimplicial map $f_{**}: T_{**} \to S_{**}$, if for all $q$:
    \begin{align*} |p \to f_{p,q}|: |p \to T_{p,q}| \to |p \to S_{p,q}| \end{align*}
    is a homotopy equivalence then $|\text{diag}f|$ is as well.

Given these two facts we can transform our original problem into something that is more readily proven. Let $T_{**} = \sqcup_*N\tilde{F}$ and $S_{**} = (N\int_J F)_**$. What (1) tells us is that we have a natural isomorphism:

\begin{align*} |p \to \text{diag}\sqcup_*N\tilde{F}(p,p)| \cong |q \to |p \to \sqcup_*N\tilde{F}(p,q)|| \end{align*}

What $(2)$ tells us is that if we take an arbitrary $q$ and show that $\Lambda(*,q): \sqcup_*N\tilde{F}(*,q) \to (N\int_J F)_{*q}$ is a homotopy equivalence than $|\text{diag}(\Lambda) = \lambda_2$ will be a homotopy equivalence. Then, note that $\Lambda(*,q)$ is the coproduct over all $q-$simplices $(L_0, X_0) \leftarrow \dots \leftarrow (L_q,X_q)$ of $N\int_J F$ of the map

\begin{align*} |p \to \{(J_0 \leftarrow J_1 \leftarrow \dots \leftarrow J_p, J_p \leftarrow L_0)\}| \to |p \to \{*\}| = \{*\} \end{align*}

We also know that $\Lambda(*,q)$ is the classifying space of the category $L_0 \backslash \mathsf{J}$(this is defined to be geometric realization composed with the nerve functor). It is contractible which implies that it is a homotopy equivalence. The last step is to show that there does exist a simplicial homotopy $H: (\text{hocolim}N\tilde{F}) \times \Delta[1] \to N\int_J F$ from $\eta \lambda_1 \to \lambda_2$. Here $\Delta[1]$ is the nerve of the category with objects $0,1$ and one non-identity map $1 \to 0$. We first want to se what a $p-$simplex of $\text{hocolim} N\tilde{F} \times \Delta[1]$ looks like. We know what a $p-$simplex in $N\tilde{F}$ looks like, so now we have to think about what it would look like in $\Delta[1]$. Turns out, it is kind of what you expect so the $p-$simplex is given by

\begin{align*} (J_0 \leftarrow_{j_1} \dots \leftarrow_{j_p} J_p, J_p \leftarrow_l L_0, (L_0, X_0) \leftarrow \dots \leftarrow (L_p, X_p)) \times (0 \leftarrow 0 \leftarrow \dots \leftarrow 0 \leftarrow 1 \leftarrow \dots \leftarrow 1) \end{align*}

where there are $i$ 0s and $p+1 - i$ 1s with $0 \leq i \leq p+1$. Now, we want to define what the homotopy $H$ actually does. $H$ will send a $p-$simplex to the following(do bare with me it gets messy):

\begin{align*} (J_0, F(j_1 \dots j_pl)(X_0)) &\leftarrow_{(J_1, F(j_1\dots j_pl)(u_1))} (J_1, F(j_2\dots j_pls_1)(X_1)) \leftarrow \dots \\ & \leftarrow (J_{i-1}, F(j_{i-1} \dots s_{i-1})(X_{i-1})) \leftarrow_{j_i\dots s_, F(j_i\dots s_{i-1})(u_i)} (L_i, X_i)\leftarrow \dots \\ & \leftarrow(L_p, X_p)\end{align*}

Finally, the proof is complete! I'm thinking of adding a more visual version of this proof to the page but for now this is all I got.