Why can I replace this direct sum isomorphism with an equality? 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.
 A: Let's recall few definitions:


*

*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.

*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 $G=A\oplus B$. 

*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. 
