Dtyjlui

The following is a question from Hatcher's "Algebraic Topology"

“show that homology commutes with direct limits. “

I have tried to solve this problem but I can’t .

It's not true in general : let $A_n subset S^1$ be the closed arc that is the image of $[frac{1}{n+1}, 1-frac{1}{n+1}]$ under $xmapsto e^{2ipi x}$ .

Let $Z(n) = S^1/A_n$ with $z_n : Z(n)to Z(n+1)$ the obvious quotient map. This provides a directed system $mathbb{N}to mathbf{Top}$.

Now $Z(n) cong S^1$ and the quotient map $Z(n)to Z(n+1)$ is the quotient by a contractible sub-CW-complex, so it's a homotopy equivalence, hence induces isomorphisms on homology.

Thus $varinjlim_n H_p(Z(n), R) = varinjlim_n H_p(S^1,R)$ where the maps are the identity, so $varinjlim_n H_p(Z(n), R) = H_p(S^1,R)$.

Let's now compute $varinjlim_n Z(n)$. I won't expand on the intuition but you can try to see it for yourself: the way $Z(n)$ varies with $n$ sort of tells us that the direct limit should be the Sierpinski space $mathbb{S}$, i.e. a space with $2$ points $a,b$ with $a$ open and $b$ closed.

To prove that, define $p_n :Z(n)to mathbb{S}$ by sending everything to $a$ except for $1in S^1$ which is sent to $b$. This is indeed a continuous map because $p_n^{-1}({a}) = Z(n)setminus{1}$ which is open. Moreover, we clearly have $p_{n+1}circ z_n = p_n$.

Now suppose we have a system of continuous maps $f_n : Z(n) to X$ such that $f_{n+1}circ z_n = f_n$. Then consider the induced maps $f_ncirc pi_n : S^1to X$ (where $pi_n : S^1to Z(n)$ is the projection). Then $f_ncirc pi_n$ sends $A_n$ to a single point then to $a$.

But $f_ncirc pi_n = f_{n+1}circ z_n circ pi_n = f_{n+1}circ pi_{n+1}$ so these induced maps are all the same (you should draw diagrams to make this clearer). Let's call it $g: S^1to X$. Then $g(A_n) subset {a}$ by what I said above, so that $g(S^1setminus{1}) = g(displaystylebigcup_n A_n)subset {a}$. Moreover, clearly $g(1) = b$. From this it's clear that the $f_n$ factor through the $p_n$.

Thus $varinjlim_n Z(n) = mathbb{S}$. But now $mathbb{S}$ is contractible, so $H_1(mathbb{S},R) = 0 neq R = H_1(S^1,R)$. Therefore homology doesn't commute with direct limits in general.

However if you start from a directed system, say indexed by $mathbb{N}$ such that each $X(n)to X(n+1)$ is the inclusion of a $T_1$ subspace, then for all $p$, $H_p(varinjlim_n X(n),R) = varinjlim_n H_p(X(n),R)$. This is true essentially because of the following lemma :

With the hypotheses above, if $Ksubset varinjlim_n X(n)$ is compact then for some $n$, $Ksubset X(n)$.

This is not trivial, but here you can find a proof of it.

Then the result will follow from simple considerations on simplices $sigma : |Delta^n|to X(infty) = varinjlim_n X(n)$: if you have $xin C_n(X(infty),R)$ (the singular chains), then it's actually in $C_n(X(m),R)$ for some $m$, and it's a cycle in one if and only if it's a cycle in the other and it's a boundary in $C_n(X(infty),R)$ if and only if for some $pgeq m$, it's a boundary in $C_n(X(p),R)$, so that $Z_n(X(infty),R) = varinjlim_m Z_n(X(m),R), B_n(X(infty),R) = varinjlim_m B_n(X(m),R)$; and then the result will follow by abstract nonsense (or you can check it by hand) : $H_n(X(infty), R)= varinjlim_m H_n(X(m), R)$

Here is another easy counterexample: take $S^1$ and consider ${S_isubset S^1:S_i text {is countable}}.$ Then, $S_i$ is totally disconnected, so $H_1(S_i)=0$. And $varinjlim S_i=S$ but $H_1(S)=mathbb Z$.