3
$\begingroup$

One of the assignments for my algebraic topology course is to solve the following problem from Topology and Geometry by Bredon.

Multiplication by the prime $p$ fits a short exact sequence $$0\to \mathbb{Z}\stackrel{p}{\to}\mathbb{Z}\to \mathbb{Z}_p\to 0.$$ Use this to derive the natural split exact sequence $$0\to \frac{H_n(X)}{p H_n(X)}\to H_n(X; \mathbb{Z}_p) \to \ker\{ p: H_{n-1}(X)\to H_{n-1}(X)\}\to 0$$ (The splitting is not natural).

Where $H_n(X; \mathbb{Z}_p)$ is the homology of the chain complex $\Delta_*(X)\otimes \mathbb{Z}_p$.

I have derived the short exact sequence as follows. A short exact sequence of the form given induces a short exact sequence of chain complexes: $$0\to \Delta_*(X)\otimes \mathbb{Z}\stackrel{1\otimes p}\to \Delta_*(X)\otimes \mathbb{Z}\stackrel{1\otimes \pi}\to \Delta_*(X)\otimes \mathbb{Z}_p\to 0$$ Using the natural isomorphism $\Delta_*(X)\otimes \mathbb{Z}\cong \Delta_*(X)$ we have the following short exact sequence of chain complexes:$$0\to \Delta_*(X)\stackrel{p}\to\Delta_*(X)\stackrel{\pi}\to \Delta_*(X)\otimes \mathbb{Z}_p\to 0$$ We recall that such a short exact sequence induces a long exact sequence of homology groups: $$\cdots\to H_{n}(X)\stackrel{p_*}{\to} H_n(X)\stackrel{\pi_*}{\to} H_n(X; \mathbb{Z}_p)\stackrel{\partial_*}{\to} H_{n-1}(X)\stackrel{p_*}{\to} H_{n-1}(X)\to \cdots$$ With $\partial_*[[c]]:= [p_*^{-1}\circ \partial\circ \pi^{-1}(c)]]$ for all $c\in Z_n(\Delta_*(X)\otimes \mathbb{Z}_p)$. We can construct the following sequence $$0\to \mathrm{im}(\pi_*)\stackrel{\iota}{\to}H_n(X;\mathbb{Z_p})\stackrel{\partial_*}{\to}\ker \{p:H_{n-1}(X)\to H_{n-1}(X)\}\to 0$$ and $\textrm{im}(\pi_*)\cong \frac{H_n(X)}{\ker \pi_*}=\frac{H_n(X)}{\textrm{im}(p_*)}=\frac{H_n(X)}{pH_n(X)}$ by the first isomorphism theorem and exactness. Exactness of this sequence follows from the exactness of the long exact sequence. This gives us the desired short exact sequence $$0\to \frac{H_n(X)}{pH_n(X)}\to H_n(X;\mathbb{Z}_p)\to \ker\{p: H_n(X)\to H_n(X)\}\to 0$$

I have derived the short exact sequence, but how can I find a splitting? Is there any intuition that might lead me in the right direction?

$\endgroup$

1 Answer 1

2
$\begingroup$

Your derivation of the short exact sequence is correct. The cop-out answer to why the sequence is split is to note that it is actually not just a sequence of $\mathbb{Z}$-modules, but a sequence of $\mathbb{Z}_p$-modules, which is split by linear algebra, because $\mathbb{Z}_p$ is a field. This might be useful to consider, but it's not quite enlightening, so let's try constructing an explicit splitting.

I don't think there's too much of an intuition besides simply wanting to "reverse" the diagram chase we do in order to construct the boundary map $H_n(X;\mathbb{Z}_p)\rightarrow H_{n-1}(X)[p]$ (this notation just means $p$-torsion, but it's shorter). So how does this map work? A homology class in $H_n(X;\mathbb{Z}_p)$ is represented by a cycle in $\Delta_n(X)\otimes\mathbb{Z}_p$, which we can lift to a chain in $\Delta_n(X)$. The differential of this chain is a chain in $\Delta_{n-1}(X)$, which is divisible by $p$, and dividing it by $p$ yields a cycle representing the desired homology class in $H_{n-1}(X)[p]$. So, what can go wrong if we try to do it the other way round? Start with a homology class $[\alpha]\in H_{n-1}(X)[p]$. This means $[p\alpha]=p[\alpha]=0$, so we can find a chain $c\in\Delta_n(X)$, such that $\partial_nc=p\alpha$. Then, we can simply take $c\mod p$ and this defines a homology class in $H_n(X;\mathbb{Z}_p)$, because $\partial(c\mod p)=\partial(c)\mod p=p\alpha\mod p=0$. You can check that $[c\mod p]$ is independent of the choice of representative $\alpha\in Z_{n-1}(X)$, but it does depend on the choice of $c$.

The solution to this conundrum is that we can find a right-inverse to the differential $\partial_n\colon\partial_n^{-1}(p\Delta_{n-1}(X))\rightarrow p\Delta_{n-1}(X)\cap B_{n-1}(X)$ (since the codomain is free), call it $s\colon p\Delta_{n-1}(X)\cap B_{n-1}(X)\rightarrow\partial_n^{-1}(p\Delta_{n-1}(X))$ (note that domain and codomain are precisely the two submodules we passed through in our incomplete sketch). Now, I leave it to you to check that $H_{n-1}(X)[p]\rightarrow H_n(X;\mathbb{Z}_p),\,[\alpha]\mapsto[s(p\alpha)\mod p]$ is well-defined and a right-inverse to the differential $H_n(X;\mathbb{Z}_p)\rightarrow H_{n-1}(X)[p]$ (note that it's a right-inverse, but usually not a left-inverse, precisely because $s$ is right-inverse to $\partial_n$, but not left-inverse).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .