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Given any continuous map between topological spaces $f: X \rightarrow Y$, we can define the mapping cone $C_{f}:=((X \times[0,1]) \sqcup Y) / \sim,$ with the equivalence relation generated by $(x, 0) \sim f(x)$ for any $x \in X$ and $\left(x_{1}, 1\right) \sim\left(x_{2}, 1\right)$ for any $x_{1}, x_{2} \in X .$ Now let $f: X \rightarrow Y$ and $g: X \rightarrow Y$ be two homotopic maps. Show that then, the mapping cones $C_{f}$ and $C_{g}$ are homotopy equivalent.

I’ve made a search and have seen a few other similar questions. However, I’d like to ask some questions below, and hopefully can come up with the solution myself.

Here’s the picture of $C_f$ and $C_g$. My task would be to find maps $\alpha, \beta$ such that $\beta \circ \alpha \simeq id_{C_f}$ and $\alpha \circ \beta \simeq id_{C_g}$.

enter image description here

Say we want to define $\alpha$ first. My guess is that there are three kinds of point that I need to consider:
$\bullet p \in X \times (0,1]$
$\bullet p \in Y \setminus (f(X) \cup g(X))$, and
$\bullet p \in X \times \{0\}$

I’m thinking of letting $\alpha(p) := p$ for the first two kinds. Doing the same for $\beta$ and we’d have $\beta \circ \alpha = id_{C_f}$ and $\alpha \circ \beta = id_{C_g}$, with the domains restricted to the first two kinds of points. As for the third kind, I’m not really sure. What would happen if I also use the identity map there? If that doesn’t work, any hints on how to use the homotopy between $f$ and $g$ here?

ensbana
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    You must involve a homotopy between $f$ and $g$. It means you can continuously transform $f$ to $g$ in the 'plane' of $Y$, and vice-versa, then just apply the same deformation on the cones. – Berci Jan 15 '21 at 21:56
  • Ok, sorry. To your question: yes, you can use the identity on the whole $Y$ and only involve the homotopies to obtain the homotopies of the cones. – Berci Jan 15 '21 at 23:46
  • I’m getting confused with the what’s happening in $X \times {0}$. What would go wrong with setting $\alpha: C_f|{X \times {0}} \to C_g|{X \times {0}}$ to be $\alpha(p) = p$ and $\beta: C_g|{X \times {0}} \to C_f|{X \times {0}}$ to be $\beta(p) = p$ (so that we even have $\alpha \circ \beta = id_{C_f}$ and $\beta \circ \alpha = id_{C_g}$), and so we don’t even need the homotopy between $f$ and $g$? Also, a point $(p,0)$ would be identified with $f(p) \in C_f$ and with $g(p) \in C_g$. How would $\alpha$ and $\beta$ work on these points? – ensbana Jan 16 '21 at 11:11

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