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I am trying to answer an exercise of Hatcher's "Algebraic Topology", Section $4$.F, exercise $3$. Suppose we are given a sequence of pointed topological spaces : $Z_0\rightarrow Z_1\rightarrow Z_2 \rightarrow \cdots$. We are now interested with its homotopy colimit (i.e. what Hatcher calls "its mapping telescope"), and the behavior of based loopspaces $\Omega$.
We need to prove that there is a natural map : $\textbf{hocolim}_n \Omega Z_n \longrightarrow \Omega (\textbf{hocolim}_n Z_n)$ which is a weak equivalence.

I'm having already troubles to define the natural map, and i have no idea how to proceed for the weak equivalence proof. I know there is some kind of a filtered colimit argument, but i really want to prove this "by hand". Any ideas ?

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2 Answers 2

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First this is not true if we don't suppose some kind of separation axioms on the spaces $Z_n$. So let us suppose that $Z_n$ are all $T_1$-spaces. To be completely rigorous, i will remind the various definitions. Let us denote by $Z_0\stackrel{f_0}\rightarrow Z_1\stackrel{f_1}\rightarrow Z_2 \rightarrow \cdots$ the given sequence.

$\bullet$ Recall that $\textbf{hocolim}_n\Omega Z_n$ is obtained the following way. We have the induced sequence of based spaces : $\Omega Z_0\stackrel{\Omega f_0}\rightarrow \Omega Z_1\stackrel{\Omega f_1}\rightarrow \Omega Z_2\rightarrow \cdots$. Define the mapping cylinder $M_{n+1}$, for all $n\geq 0$, as the pushout : $$ \require{AMScd} \begin{CD} \Omega Z_n @>>> \Omega Z_n\times [n,n+1]\\ @V{\Omega f_n}VV @VVV \\ \Omega Z_{n+1} @>>> M_{n+1}. \end{CD} $$ Define $Y_0=\Omega Z_0\times \{ 0\}$, and $Y_1=M_1$. For all $n\geq 1$, define : $$Y_{n+1}=M_1\coprod_{\Omega Z_1\times \{ 1\}} M_2 \coprod_{\Omega Z_2\times \{2\}} \cdots \coprod_{\Omega Z_{n-1}\times \{n-1\}}M_n.$$ Notice that we have the homotopy equivalences $r_n:Y_n\stackrel{\simeq}\rightarrow \Omega Z_n$, such that the following diagram commutes for all $n\geq 0$ : $$ \require{AMScd} \begin{CD} \Omega Z_n @<{\simeq}<{r_n}< Y_n\\ @V{\Omega f_n}VV @VV{inclusion}V \\ \Omega Z_{n+1} @<{\simeq}<{r_{n+1}}< Y_{n+1}. \end{CD} $$ Then $\textbf{hocolim}_n\Omega Z_n=\textbf{colim}_nY_n=\left(\bigcup_{n\geq 0}\Omega Z_n\times [n,n+1]\right)/\sim$, where we identified $(\lambda_n, n+1)$ in $\Omega Z_n\times [n,n+1]$ with $(\Omega f_n(\lambda_n), n+1)$, but seen in $\Omega Z_{n+1}\times [n+1, n+2]$.

$\bullet$ Similarly we define $\textbf{hocolim}_nZ_n$ as $\textbf{colim}_nW_n$, where $W_0=Z_0\times \{ 0\}$, $W_1=N_1$ and for all $n\geq 1$ : $$W_{n+1}=N_1\coprod_{Z_1\times \{ 1\}} N_2 \coprod_{Z_2\times \{2\}} \cdots \coprod_{Z_{n-1}\times \{n-1\}}N_n,$$ where $N_{n+1}$ is the mapping cylinder defined as the pushout for $n\geq 0$ : $$ \require{AMScd} \begin{CD} Z_n @>>> Z_n\times [n,n+1]\\ @V{ f_n}VV @VVV \\ Z_{n+1} @>>> N_{n+1}. \end{CD} $$ We also have homotopy equivalences $s_n:W_n\stackrel{\simeq}\rightarrow Z_n$ which commute with the inclusions $W_n\rightarrow W_{n+1}$.

$\bullet$ Now let us define the natural map : $G:\textbf{hocolim}_n\Omega Z_n\rightarrow \Omega\textbf{hocolim}_n Z_n$. As suggested by Brian Rushton, we define $G$ the following way. Let us define $g:\bigcup_{n\geq 0}\Omega Z_n\times [n,n+1]\rightarrow \Omega\textbf{hocolim}_n Z_n$ by assigning each $(\lambda_n, m)$ in $\Omega Z_n\times [n, n+1]$ a map : $$I\rightarrow Z_n\times \{ m\} \stackrel{inclusion}\rightarrow Z_n\times [n,n+1] \stackrel{inclusion}\rightarrow \bigcup_{n\geq 0} Z_n\times [n,n+1]\stackrel{quotient}\rightarrow \textbf{colim}_nW_n,$$ by $t\mapsto (\lambda_n(t),m)$. This defines our map $g$ (obviously continuous using the exponential law). Of course : $g(\lambda_n, n+1)=g(\Omega f_n(\lambda_n),n+1)$ by the identifications on $\textbf{colim}_nW_n$. Then we obtain our unique map $G$ by the property of quotients.

$\bullet$ Now for every $m\geq 0$ we have : $$ \begin{array}{llll} \pi_m(\textbf{hocolim}_n\Omega_nZ_n) & = & \pi_m(\textbf{colim}_nY_n) & \\ & \cong & \textbf{colim}_n\left(\pi_m(Y_n)\right) & \mbox{Here i am using that $Z_n$ are $T_1$ spaces}\\ & \cong & \textbf{colim}_n\left(\pi_m(\Omega Z_n)\right) & \mbox{Using the homotopy equivalences $r_n$}\\ & \cong & \textbf{colim}_n\left(\pi_{m+1}(Z_n)\right) &\\ & \cong & \textbf{colim}_n\left(\pi_{m+1}(W_n)\right) & \mbox{Using the homotopy equivalences $s_n$}\\ & \cong & \pi_{m+1}\left(\textbf{colim}_nW_n\right) & \\ & \cong & \pi_m\left(\Omega \textbf{hocolim}_nZ_n\right). & \end{array} $$ These isomorphisms are exactly the map $G_*:\pi_m(\textbf{hocolim}_n\Omega_nZ_n)\rightarrow \pi_m\left(\Omega \textbf{hocolim}_nZ_n\right)$. Therefore $G$ is a weak equivalence.

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Hi mph, I think you can actually remove the separation axiom and your argument still works. This is because your mapping telescopes $Y_n$ have a natural projection into $\mathbb R$, and you can use that projection to argue that $\mathbf{colim}_n(\pi_m Y_n) \rightarrow \pi_m (\mathbf{colim}_n Y_n)$ is an isomorphism. The separation axiom is still important if you don't use a mapping-telescope construction and instead build the hocolims by taking an honest colim along a sequence of cofibrations. –  Cary Feb 8 at 17:00
    
@Cary Thank you! You are absolutely right! I shall edit when I will have time. –  mph Feb 15 at 11:34

For the natural map, each point on the left lies in some loopspace cross an interval, so is represented by a map from the circle into one of the $Z_i$ together with a number $t$ between 0 and 1; the map from the circle induces a map from the circle the telescope (the telescope contains many copies of $Z_i$, and we use the one indexed by $t$); this map from the circle gives a point in the loopspace.

Now this might not be well defined because some points are in two loopspaces, inducing two maps into two different $Z_i$, but these two maps become the same in the telescope (this always happens when $t=0$ in one loopspace and 1 in another; but our map sends such things to the same place.).

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It could just be the typos, but I don't understand the last paragraph. Could you add some more detail please? –  Daniel Rust Dec 21 '13 at 16:35
    
@DanielRust I tried again, but it may need more work. –  Brian Rushton Dec 22 '13 at 3:14
1  
Ok i have figured out the answer, thank you Brian Rushton! I will now write the answer. –  mph Dec 26 '13 at 22:59

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