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In response to the following question from Hatcher's Algebraic Topology:

Construct explicit chain maps $s : C_n(X) \to C_{n+1}(SX)$ inducing isomorphisms $\tilde{H_n}(X)\simeq \tilde{H}_{n+1}(SX)$.

one possible solution that was presented (here at the end of 2.1.21; it also appears here) is the following:

...construct $s$ straight from the difference between the embedded north cone and the south cone, say $s(\sigma) = S\sigma_{north} - S\sigma_{south}$ where $S$ is the suspension operator.

First, why not just take the suspension as is? i.e.: $s(\sigma) = S\sigma$.

Second, why the minus? I can tell it's probably related to the signs when taking the boundaries somehow, but I'm unable to see it formally when trying to spell out the boundary or when drawing an example.

Any help would be much appreciated.

Anon
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    When you say that $S$ should induce a chain map because it is continuous, what do you mean? $S$ is not a map between spaces but an operation on spaces. It is true that there is a map $X\to SX$ given by including $X$ into the equatorial slice of the suspension, but this map induces the trivial map on homology. – Cheerful Parsnip Feb 12 '23 at 19:11
  • @CheerfulParsnip Yes - that's what I meant. Just to make sure I understand why it's the trivial map on homology - that's because the map is to the $n+1$-dimensional homology group, where $i(X)$ (the inclusion's image) would have no $n+1$-simplices - hence the trivial homology group. Correct? – Anon Feb 12 '23 at 19:24
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    Yes, that explains why it is has trivial image in degree $n+1$, but in fact it has trivial image in all degrees except $0$. This is because the image $i(X)$ is null homotopic in $SX$ by shrinking it up to one of the two cone points. – Cheerful Parsnip Feb 12 '23 at 23:03

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