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Let $f,g:X\to Y$ be maps of spaces such that $f\simeq g$. Is it true that the mapping cones $\operatorname{cone}(f)$ and $\operatorname{cone}(g)$ are homotopy equivalent?

Can we write down an explicit homotopy equivalence? It ought not to be too hard to get a map $\operatorname{cone}(f)\to \operatorname{cone}(g)$ from the existence of a homotopy $H:X\times I\to Y$, but I haven't been able to find a sensible candidate, despite trying for quite a while.

John Steinbeck
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1 Answers1

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Define the map $k: \text{cone}(f) \to \text{cone}(g)$ by $$ \begin{align} % Y \sqcup \left(X\times\left[0,\frac12\right]\right) \sqcup \left(X\times\left[\frac12,1\right]\right) & \to Y \sqcup \left(X\times I\right) \\ y & \mapsto y \qquad \text{ for } y\in Y \\ (x,t) & \mapsto \begin{cases} H(x,2t) &\text{ if } t\le\frac12\\ (x,2t-1) &\text{ if } t\ge\frac12 \end{cases} \end{align} $$ Can you show that $k$ is continuous?

Now define the "inverse" $l: \text{cone}(g) \to \text{cone}(f)$ $$ \begin{align} y & \mapsto y \qquad \text{ for } y\in Y \\ (x,t) & \mapsto \begin{cases} H(x,1-2t) &\text{ if } t\le\frac12\\ (x,2t-1) &\text{ if } t\ge\frac12 \end{cases} \end{align} $$ Let $m: \text{cone}(f) \to \text{cone}(f)$ be the map which is the identity on $Y$ and which $$ (x,t) \mapsto \begin{cases} f(x) &\text{ if } t\le\frac34\\ (x,4t-3) &\text{ if } t\ge\frac34 \end{cases} $$

A homotopy between $lk$ and the map $m$ is given by $$ (x,t,s) \mapsto \begin{cases} H(x,2t(1-s)) &\text{ if } t\le\frac12\\ H(x,(3-4t)(1-s)) &\text{ if } \frac12\le t\le\frac34 \\ (x,4t-3) &\text{ if } t\ge\frac34 \\ \end{cases} $$ Can you show that $m$ is homotopic to the identity on $\text{cone}(f)$?

Stefan Hamcke
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    How would you think of coming up with something like this? Why, if I'm told to prove this result, would I not try to come up with a homotopy between $lk$ and the identity map? How to visualise the motivation for coming up with this? – DpS Mar 01 '18 at 22:32
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    Also seems to me like $k$ is mapping $(x,t)$ to two different elements, namely $g(x)$ and $[(x,0)]$, when $t = 1/2$. – DpS Mar 02 '18 at 15:28
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    @DpS Let me know if your definition is different, but we defined the mapping cone of $f$ to be $[(X \times [0, 1]) \sqcup Y] / \sim$, where $\sim$ is the equivalence relation generated by $(x, 0) \sim f(x)$ and $(x, 1) \sim (\bar{x}, 1)$ for all $x, \bar{x} \in X$. – ViktorStein Nov 26 '20 at 23:48
  • Can I get some help with showing $k$ is continuous? Here’s my attempt so far. For $y \in Y \subseteq C_f$, $k(y)$ is the identity function, hence continuous. For $(x,t) \in C_f$ with $t \leq 1/2$, $k$ is the product of two continuous functions $x \mapsto x$ and $t \mapsto 2t -1$, hence continuous. – ensbana Jan 30 '21 at 16:28
  • I’m not entirely sure about the remaining case. I think the continuity of $H$ should be enough to make $k$ continuous when $t \geq 1/2$. But here the image of $(x,t)$ under $k$ is in $(X \times {0})$, whereas the co-domain of $H$ is $Y$. How do I use the equivalence relation $(x,0) \sim g(x)$ to make this more rigorous? – ensbana Jan 30 '21 at 16:28
  • Perhaps to make things explicit, for $t \leq 1/2$, this is what happens with $k$?

    $C_f|_{X \times [0,1/2]} \xrightarrow{(x,t) \ \mapsto \ (x,2t)} X \times I \xrightarrow{H} Y \hookrightarrow C_g$, where $\hookrightarrow$ is the inclusion function.

     – ensbana Jan 30 '21 at 17:28
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    @ensbana Yes, the composition of maps in your last comment shows that $k$ is continuous on $X×[0,1/2]$. Regarding your concern about the case $t=1/2$, this can either be seen as being mapped to $H(x,1)$, or as being mapped to $(x,0)$. But that is not a problem, since $H(x,1)$ equals $g(x)$, and that point is identified with $(x,0)$, thus is the same as $(x,0)$ in $C_g$. – Stefan Hamcke Feb 04 '21 at 08:15