1

Let $\phi_{n,k}$ denote $\Sigma^{n-1} \phi_k : S^n \longrightarrow S^n.$ Compute $H_*(C(\phi_{n,k})),$ where $C(\phi)$ is the mapping cone of $\phi.$

Here $\Sigma^{n-1}$ denotes the $(n-1)$-fold reduced suspension and $\phi_k$ denotes multiplication by $k.$

How do I compute it? It turns out to me that the problem can be done using Mayer-Vietoris long exact sequence if it is known what's the effect of $\Sigma^{n-1} \phi_k$ in the level of homology. But I am unable to figure out the induced homomorphism. Any help in this regard would be much appreciated.

Thanks in advance.

Fanatics
  • 211
  • 1
    Are you familiar with the result that suspension is a natural isomorphism? – Elliot Yu Jul 01 '21 at 16:10
  • @Elliot Yu I am not familiar with that. – Fanatics Jul 01 '21 at 16:25
  • 1
    The result essentially says that the suspension isomorphism commutes with homomorphisms induced by a map between two spaces. The answer in the question I linked to contains an outline of a proof which relies on the naturality of the long exact sequence and excision. I'm not a hundred percent sure about how to prove the naturality of excision (or whether it's truly necessary). In any case, if you accept this result, then you can figure out the effect of $\Sigma^{n-1}\phi_k$ at degree $m$ by looking at the effect of $\phi_k$ at degree $m - n + 1$. – Elliot Yu Jul 02 '21 at 17:46
  • 2
    Also just as a small matter of typesetting, may I suggest using \Sigma instead of \sum to write the symbol for suspension? The former generates an upper case sigma that is treated the same way as upper case Latin letters, while the latter generates a symbol with variable size, and when used in display style equations (i.e. when you put the expressions in $$ $$), the superscript gets moved to the top of the \sum. – Elliot Yu Jul 02 '21 at 17:54

0 Answers0