Why can I replace this direct sum isomorphism with an equality

algebraic-topologyhomology-cohomologyproof-explanation

From Rotman's Algebraic Topology

Theorem: For all $n \ge 0, $ $$\tilde H_n(X) \approx H_n(\tilde S_*(X), \partial)$$ where $\tilde H$ means reduced homology groups.

Proof: There's a short exact sequence $0 \rightarrow \text {ker } \tilde \partial_0 \rightarrow S_0(X) \xrightarrow{\tilde \partial_0} \tilde S_{-1}(X) \rightarrow 0$. If $\alpha \in S_0(X)$ satisfies $\tilde \partial_0(\alpha) = 1,$ then it is easy to see that $S_0(X) = \text{ker } \tilde \partial_0 \oplus \langle \alpha \rangle$ and $\langle \alpha \rangle \approx \Bbb Z.$ But $\tilde \partial_0 \partial_1 = 0 $ implies that $B_0(X) = \text{im } \partial_1 \subset \text{ker } \tilde \partial_0$. Since $S_0(X) = Z_0(X), $ we have:

$(*) \space H_0(X) = S_0(X)/B_0(X) = (\text{ker } \tilde \partial_0 \oplus \langle \alpha \rangle) / B_0(X) $ $\approx (\text{ker }\tilde \partial _0 / B_0(X)) \oplus \Bbb Z = \tilde H_0(X) \oplus \Bbb Z.$

The author then footnotes:

This is a special case of a more general result: If $0 \rightarrow K \rightarrow G \rightarrow F \rightarrow 0$ is exact with $(K \rightarrow G)$ an inclusion with $F$ free abelian, then $G = K \oplus F'$, where $F' \approx F$. Here we present a proof of this special case. If $x \in \text{ker} \tilde \partial_0 \cap \langle \alpha \rangle,$ then $x = m\alpha$ and $\tilde \partial_0(x) = 0 = m$, hence $x=0$; if $\gamma \in S_0(X),$ then $\tilde \partial_0(\gamma)= k$, say, and so $\gamma = (\gamma – k \alpha) + k \alpha \in \text{ker } \tilde \partial_0 + \langle \alpha \rangle$.

But I'm a little confused on the notation here. I see that in the footnote the author shows that $S_0(X) = \text{ker } \tilde \partial_0 + \langle \alpha \rangle \approx \text{ker } \tilde \partial_0 \oplus \langle \alpha \rangle$, which implies that $S_0(X) \approx \text{ker } \tilde \partial_0 \oplus \langle \alpha \rangle$. But in the proof of the theorem the author states that explicitly $S_0(X)$ is exactly equal to $\text{ker } \tilde \partial_0 \oplus \langle \alpha \rangle$.

What exactly am I missing here? The footnote seems to indicate a direct sum, along with the line in the partial proof. But the proof in the footnote seems to only prove that it's they're isomorphic and not equal.

Best Answer

Let's recall few definitions:

  1. If $A,B\subseteq G$ are subgroups (more generally submodules) then $A+B=\{a+b\ |\ a\in A, b\in B\}$. This is a subgroup (submodule) if $G$ is abelian. This is not the internal direct product/sum. It is just a sum of subgroups.
  2. If $A,B\subseteq G$ are subgroups of an abelian group $G$ then we say that $G$ is the internal direct product/sum of $A$ and $B$ if $G=A+B$ and $A\cap B=0$. In this case we write $G=A\oplus B$.
  3. If $A,B$ are (arbitrary) groups then their external direct product/sum is defined as the Cartesian product $A\times B$ as a set with pointwise group addition $(a,b)+(a',b'):=(a+a', b+b')$. In this case we also write $A\oplus B$.

All of that is the standard notation. And there's a theorem that says that if $G=A\oplus B$ is the internal direct product then $G\simeq A\oplus B$ as the external direct product. And vice versa, the external direct product $A\oplus B$ induces internal by taking $A\times 0$ and $0\times B$.

Clearly the author uses both 1. and 2. in the footnote and $S_0(X)=\ker\tilde\partial_0\oplus \langle \alpha\rangle$ part. He then switches to the external direct product with $(\ker\tilde\partial_0\oplus \langle \alpha\rangle)/B_0(X)\simeq (\ker\tilde\partial_0/B_0(X))\oplus \langle \alpha\rangle$ isomorphism: he uses internal on the left side and external on the right side.

I agree that this is confusing, he is juggling with both terms. But since internal and external direct products are pretty much the same thing I suppose we can forgive him this nuance.

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