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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!

Skalf
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  • I would not expect this to have a nice topological solution merely based on its location in the text. That portion of Hatcher is dealing with, among other things, the algebra of exact sequences, and some problems will have purely algebraic solutions. I guess I'm quibbling with your use of the word "so" in the first sentence of paragraph 2. – John Palmieri Apr 28 '22 at 22:25
  • You are probably right, I would not be surprised at all if the author expected a purely algebraic proof here. Allow me to rephrase this weird "so". Regardless of the context, I would be curious about a topological approach to this exercise, even if it is harder. Now, of all the various ways one can think of to solve this, the location of the exercise in the book made me think first about long exact sequences in homology, simply because this is the last thing I have seen. It might not be the correct approach! – Skalf Apr 29 '22 at 07:26
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    Here is a comment that not intended to address your question, but is nonetheless intended to challenge the premise behind the question; I hope you don't mind. The whole purpose of algebraic topology is to translate hard-to-solve problems of topology into easy-to-solve problems of algebra. The highlighted exercise is, from the point of algebra, intended as an easy exercise about some of the simplest groups in group theory. So if you eventually wish to apply algebraic topology tools, you'll need some algebra skills. – Lee Mosher Apr 29 '22 at 14:51

1 Answers1

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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 enter image description here

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 enter image description here

Amadeus Maldonado
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  • Crystal clear, thank you very much! Even better, your construction only involves cells of dimension 2 or less (these spaces are of special interest to me) and is very intuitive. I initially thought about copying the algebraic application constructed by sending 1 to (2,1) but I did not see the 8-shape at the extremity of the cylinder as a circle and I got stuck, this is a simple but smart trick – Skalf May 02 '22 at 12:34
  • Can you please explain why $\widetilde H(X)=\mathbb Z_8\times \mathbb Z_2$. – Biplab Feb 13 '24 at 06:40