Homotopic maps have homotopy equivalent mapping cones

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.

Solution 1:

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 $\text{cone}(f)$?