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I'm reading Munkre's elements of algebraic topology. Lemma 58.1 is the following:

Let $C$ and $C'$ be chain complexes such that in each dimension the cycles form a direct summand in the chains. (This occurs, for instance, when $C$ and $C'$ are free). Then:

$$\Theta: \oplus_{p+q=m} H_p(C) \otimes H_q(C') \rightarrow H_m(C \otimes C')$$

is a monomorphism, and its image is a direct summand.

On the level of chains, is this an isomorphism?

I'd appreciate any insight into this Lemma what so ever, the explanations I get from people here always really help my understanding. In particular though, what exactly does it mean for "The cycles to be a direct summand in the chains"?

Thanks!

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  • $\begingroup$ Presumably here "free" means free over a PID (so that it suffices to be free over $\mathbb{Z}$). $\endgroup$ Sep 3, 2019 at 15:12

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The cycles of $C$, $Z(C)$, is in general a subgroup of $C$. Asking for $Z(C)$ to be a direct summand is asking for the existence of another subgroup $K \subseteq C$ such that $C = Z(C) \oplus K$. This is not always true: consider the chain complex $$ \cdots \to 0 \to \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0 \to \cdots,$$ where the map $\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$ is the quotient map. Then the cycles inside of $\mathbb{Z}$ is the subgroup $2\mathbb{Z}$, which is not a summand of $\mathbb{Z}$. (Any two nonzero subgroups of $\mathbb{Z}$ intersect.)

Now on your question about an isomorphism on the level of chains. It's not clear what "chains" we would be discussing on the left-hand side: the group $$\bigoplus_{p+q=m} H_p(C) \otimes H_q(C')$$ doesn't arise as the homology of a relevant chain complex. Of course, it is the homology of a complex with zero differentials, but if $H_\ast(C)$ doesn't map into $C$, then we shouldn't expect this tensor product to map into $C \otimes C'$.


Edit: The map is induced by sending $H_p(C) \otimes H_q(C') \ni [c] \otimes [c'] \mapsto [c \otimes c'] \in H_{p+q}(C\otimes C')$. If $\varphi: C \to Z(C)$ and $\varphi':C' \to Z(C')$ are splittings, then sending

$$c \in Z(C \otimes C') \mapsto (\varphi \otimes \varphi')(c)\in Z(C) \otimes Z(C')\twoheadrightarrow H_\ast(C) \otimes H_\ast(C')$$

descends to a map on homology: the boundary map in $C \otimes C'$ is given by $\partial_C \otimes 1 \pm 1 \otimes \partial_{C'}$, and since $\varphi$ and $\varphi'$ are the identity on cycles, they are the identity on boundaries. So, a boundary in $C \otimes C'$ gets taken to zero under the above map. So we have maps

$$ \require{AMScd}\begin{CD}\bigoplus_{p+q=m} H_p(C) \otimes H_q(C') @>{[c]\otimes [c']\mapsto [c\otimes c']}>> H_m(C\otimes C') @>{\varphi\otimes \varphi'}>> \bigoplus_{p+q=m} H_p(C) \otimes H_q(C') \end{CD} $$ which compose to the identity, which shows the desired.

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  • $\begingroup$ However, isn’t his claim about free chain complexes only true if the ring is semi simple? There are obvious counter examples over the integers. $\endgroup$ Sep 3, 2019 at 14:43
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    $\begingroup$ @ConnorMalin What are those obvious counterexamples? Over a PID, if $C$ is a free chain complex, then the boundaries are free, so the short exact sequence $0 \to \ker\partial \to C \to^\partial \mathrm{im}\,\partial[-1]\to 0$ splits. $\endgroup$ Sep 3, 2019 at 15:11
  • $\begingroup$ It is my mistake. I assumed that such a thing could not have torsion in the homology, but of course that is wrong. I guess really what I was thinking is that the boundaries are not direct summands. $\endgroup$ Sep 3, 2019 at 18:13

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