algebraic topology - hatcher 3.3 exercise 17

Question

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 .

Answer 1

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 $x\mapsto e^{2i\pi x}$ .

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 $1\in S^1$ 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 $f_n\circ \pi_n : S^1\to X$ (where $\pi_n : S^1\to Z(n)$ is the projection). Then $f_n\circ \pi_n$ sends $A_n$ to a single point then to $a$.

But $f_n\circ \pi_n = f_{n+1}\circ z_n \circ \pi_n = f_{n+1}\circ \pi_{n+1}$ so these induced maps are all the same (you should draw diagrams to make this clearer). Let's call it $g: S^1\to X$. Then $g(A_n) \subset \{a\}$ by what I said above, so that $g(S^1\setminus\{1\}) = g(\displaystyle\bigcup_n A_n)\subset \{a\}$. 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 $K\subset \varinjlim_n X(n)$ is compact then for some $n$, $K\subset X(n)$.

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 $x\in C_n(X(\infty),R)$ (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 $p\geq m$, 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) : $H_n(X(\infty), R)= \varinjlim_m H_n(X(m), R)$

Answer 2

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