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Hi all,

Before asking my question, I need to fix some terms and notation.

Let $M$, $M'$ be locally compact, Hausdorff spaces, and $f:M\rightarrow M'$ a homotopy equivalence with homotopy inverse $g:M'\rightarrow M$. Set for a natural number $n$, $C^n(M):=(g\circ f)^n(M)$. Similarly, we set $C^{n}(M'):=(f\circ g)^n(M')$.

I'll call the data $(f,g,M,M')$ finitely-cored if there exist $N,N'\in \mathbb{N}$ such that $(g\circ f)$ restricted to $C^N(M)$ is the identity and $(f\circ g)$ restricted to $C^{N'}(M')$ is also the identity.

My questions are the following:

1) What are the properties of $(f,g,M,M')$ such that it is finitely-cored?

2) Under what conditions (if at all) can $(f,g)$ be deformed to another homotopy equivalence $(f_1,g_1)$ (i.e. $f\sim f_1$, $g\sim g_1$ )such that $(f_1,g_1,M,M')$ is finitely-cored?

3) For $(f,g,M,M')$ not finitely-cored, define $C_{\infty}(M):= \cap_{n=1}^\infty C^n(M)$ and similarly for $M'$. Is it true that $C_{\infty}(M)$ is homeomorphic to $C_{\infty}(M')$? If not, I'd also like to know under what conditions this holds.

Thanks in advance for your replies. Any references are also much appreciated!

P.S: I've invented the term "finitely-cored" here to make the question easier to present, so if there is already a term for this object I apologise for my ignorance.

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  • $\begingroup$ Hi Tyler, I've removed the statement, but did you remove yours too? I'm nevertheless not sure about it (I read a similar statement sommewhere and jumped the gun!) $\endgroup$
    – Indrava Roy
    Commented Mar 22, 2010 at 16:39
  • $\begingroup$ Yes, I deleted the comment that $C^{n+1}(M)$ is not necessarily a deformation retract of $C^n(M)$. $\endgroup$
    – Tyler Lawson
    Commented Mar 22, 2010 at 17:12
  • $\begingroup$ I seem to have convinced myself at one point that 1) $(g\circ f)(M)$ is a deformation retract of M , the deformation retraction being given by the homotopy between gf and the identity on M, and 2) You can repeat this argument for $C^{n+1}(M)$ and $C^n(M)$. Right now I'm a bit hazy, I need to see whether these arguments are valid. $\endgroup$
    – Indrava Roy
    Commented Mar 22, 2010 at 17:21
  • $\begingroup$ The problem is that the function $g \circ f$ is not necessarily 1-to-1, and so the homotopy between gf and the identity may not be well-defined because it depends on choosing a preimage. $\endgroup$
    – Tyler Lawson
    Commented Mar 22, 2010 at 17:41

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