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!
 A: 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.
