# What does it mean for "The cycles to be a direct summand in the chains"?

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!

• Presumably here "free" means free over a PID (so that it suffices to be free over $\mathbb{Z}$). Sep 3 '19 at 15:12

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.

• 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. Sep 3 '19 at 14:43
• @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. Sep 3 '19 at 15:11
• 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. Sep 3 '19 at 18:13