Topological proof to Hatcher's exercise 2.1.14 The first part of the exercise goes like this:

Determine whether there exists a short exact sequence $0 \rightarrow \mathbb{Z}_4 \rightarrow \mathbb{Z}_8 \oplus \mathbb{Z}_2 \rightarrow \mathbb{Z}_4 \rightarrow 0$.

It turns out the answer is yes, there does exist such a short exact sequence and I have seen several proofs of this exercise on the internet, e.g  this one or this one . However all these proofs are purely algebraic and only use group theory results: to me (I am no algebraist) this is very unsatisfying.
This exercise comes at the end of a chapter about exact sequences in (singular) homology, so I would expect that it is possible to find a topological object that gives a long exact sequence in homology containing our short exact sequence at some point. Just a few pages before the exercise, we can see that given a space $X$ (let's say it is a $\Delta$-complex) and a subspace $A \subset X$, there is a long exact sequence
$$... \quad \rightarrow H_n (A) \rightarrow H_n(X) \rightarrow H_n(X,A) \rightarrow H_{n-1}(A) \rightarrow \quad ... \quad  \rightarrow H_0(X,A) \rightarrow 0$$
Even better, when we consider reduced homology we have a long exact sequence
$$... \quad \rightarrow \tilde{H}_n (A) \rightarrow \tilde{H}_n(X) \rightarrow \tilde{H}_n(X/A) \rightarrow \tilde{H}_{n-1}(A) \rightarrow \quad ... \quad  \rightarrow \tilde{H}_0(X/A) \rightarrow 0$$
I say it is better because, provided our space $A$ is path-connected, we have $\tilde{H}_0(A)=0$ so maybe it would be possible to find spaces $X$ and $A$ such that our short exact sequence is realized by $$\tilde{H}_2(X/A) \rightarrow \tilde{H}_1(A) \rightarrow \tilde{H}_1(X) \rightarrow \tilde{H}_1(X/A) \rightarrow \tilde{H}_0(A)=0$$
and in that case it would be possible to see directly on a picture the algebraic relations between $\mathbb{Z}_4$ and $\mathbb{Z}_8\oplus \mathbb{Z}_2$, since the generators of the $H_1$'s are curves.
I tried for a bit to find such spaces $X$ and $A$, but I don't know much about homology and I severely lack examples (in fact this is the reason why I am going through these exercises) so I failed to do so. I know one can construct a space $Y$ with $H_1(Y) = \mathbb{Z}_n$ and trivial other (reduced) homology groups by glueing a $2$-cell on $\mathbb{S}^1$ with a degree $n$ map, so by taking a wedge sum of such spaces we would have a good candidate for $X$, but it is not obvious to me what I should pick for $A$. I also tried with $X$ being a lens space as described in a previous exercise (namely exercise 2.1.8) but same problem. Or the other way around, starting with $A$ I don't see how to obtain the desired $X$ by glueing extra cells to $A$.
Do you know such spaces $X$ and $A$? Any thoughts on the subject would be greatly appreciated!
 A: In your first reference for an algebraic solution, the key was simply to construct the injective map $\mathbb{Z}_4 \to \mathbb{Z}_8 \oplus \mathbb{Z}_2$ given by $1 \mapsto (2,1)$ and notice that the quotient by the image is a cyclic group of order $4$, so we'll mimic this map in our topological construction.
First pick $A$ by gluing a $2$-cell using a degree $4$ map on $\mathbb{S}^1$. Now to construct the rest of $X$, glue a cylinder to this $\mathbb{S}^1$ which glues at the other end to a figure $8$ having two cylinders coming out of it itself. These two cylinders will be the paths to transform our $1$ into a $(2,1)$ in $\mathbb{Z}_8 \oplus \mathbb{Z}_2$, therefore one of the cylinders glues to a $\mathbb{S}^1$ via a degree $2$ map and this $\mathbb{S}^1$ also has a $2$-cell glued to it via a degree $8$ map. Meanwhile, the end of the other cylinder glues to a $\mathbb{S}^1$ via a degree $1$ map and it also has a $2$ cell attached to it via  a degree $2$ map. The final picture for $X$ looks something like this

where the arrows indicate de degree of the map being used to glue each part and the red part is $A$ itself. One may check (using, for example, Mayer-Vietoris) that $\tilde{H}_1(X) = \mathbb{Z}_8 \oplus \mathbb{Z}_2$ and the map induced by the inclusion $\tilde{H}_1(A) \to \tilde{H}_1(X)$ is of the form $1 \mapsto (2,1)$.
If you'd like to go further and not use algebra to conclude that the quotient by the image is a cyclic group of order $4$, you could also directly compute $\tilde{H}_1(X/A)$, just notice that the resulting figure is a cylinder glued to spheres $\mathbb{S}^1$ and $2$-cells and use something like Mayer Vietoris

