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If $X$ is a topological space, define ${\rm Cil}(X) = X \times I$ the cylinder over $X$, and the cone over $X$, $\operatorname{Con}(X) = \operatorname{Cil}(X)/{\sim}$ the quotient by saying that $(x_1,s_1)\sim (x_2, s_2)$ iff they're equal, or if $s_1=s_2= 1$.

The exercise asks us to prove that $\operatorname{Con}(X)$ is contractible. Modulo notation, I've found a few good answers around here about this exercise (such as this and that, etc), but there's one thing I'm not getting.

In the previous exercise, I managed to prove that ${\rm Cil}(X)$ has the same homotopy type as $X$, and I did that using $H: {\rm Cil}(X) \times I \to {\rm Cil}(X)$, $H((x,s),t) = (x,st)$. I have zero experience with quotient topologies and such, but I thought of passing $H_t = H(\cdot, t)$ to the quotient somehow, as in $\widetilde{H}: {\rm Con}(X) \times I \to {\rm Con}(X)$, $\widetilde{H}(\left[(x,s)\right],t) = \left[(x,st)\right]$.

One readily sees that $\widetilde{H}_1 = {\rm id}_{{\rm Con}(X)}$ and $\widetilde{H}_0 = c_{\left[(x,0)\right]}$, so the identity is homotopic to a constant and we're done.

I understand from the answers I linked above that it is not so immediate that $\widetilde{H}$ will be continuous, but ok, I can follow what happens there.

My specific problem is more silly: seeing why $\widetilde{H}$ defined like this is actually well-defined.

Suppose that $(x_1,s_1)\sim (x_2,s_2)$. If they're equal, then $H((x_1,s_1),t) = H((x_2,s_2),t)$, ok. If they're not equal, then $s_1 = s_2 = 1$, and I'd have to check that $\left[(x_1,t)\right] = \left[(x_2,t)\right]$, but there's no reason for this to be true.

Can someone break this down for me? Thanks.

Community
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Ivo Terek
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1 Answers1

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You've done half of the homotopy with the cone upside-down. Intuitively, the homotopy equivalence should shrink the cone to its vertex, so if your equivalence relation $\sim$ identifies $(x_1,s_1)$ with $(x_2,s_2)$ when $s_1=s_2=1$, then your homotopy should be between $\mathrm{id}_{\mathrm{Con}(X)}$ and $c_{[(x,1)]}$, not between $\mathrm{id}_{\mathrm{Con}(X)}$ and $c_{[(x,0)]}$.

Had you defined $(x_1,s_1) \sim (x_2, s_2)$ iff they're equal or $s_1=s_2=0$ then I believe your original homotopy would work. (And this would yield an isomorphic cone space to the cone space you define.)

If you don't want to re-define $\sim$, then try using $$H((x,s),t) = (x,1-(1-s)t)$$ I think you'll have more success.

Clive Newstead
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  • AH. That works nicely. Very helpful. I was trying to twerk my homotopy there making $s \mapsto 1-s$ and/or $t \mapsto 1-t$, but you did $1 - $ the whole argument. Redefining $\sim$ makes the initial homotopy work, as you said, I just checked it here too. Thanks a lot, that's exactly what I needed :) – Ivo Terek Aug 24 '15 at 04:05
  • For future reference: $\widehat{H}((x,s),t) = (x, s+(1-s)t)$ also does the job ("equation of the line" passing through $(x,s)$ and $(x,1)$!) – Ivo Terek Aug 24 '15 at 04:16