Given a continuous map $f\colon X\to Y$ between two non-empty topological spaces, show that there is homotopy equivalence between the mapping cylinder $(X\times I)\sqcup _{f}Y$ and Y.
Here we have I=$[0,1]$, $ (x,1)\sim f(x)$ on X.
Is the following proof correct?:
Let's denote by $\cong$ a homotopy equivalence.
I is contractible so $I\cong \{1\}$, and obviously $X \cong X$,
So we have $X\times I \cong X\times \{1\} \cong f(X)$
Therefore $X\times I \sqcup_f Y\cong f(X)\sqcup_f Y = Y$
Thank you for your corrections and comments.
Best Answer
Denote $M :=(X\times I)\sqcup _{f}Y$ and take maps $p: Y \to M, y \mapsto y$ and $q: M \to Y, (x, s) \mapsto f(x), y \mapsto y$. Well defined: easy to verify
Then $q \circ p= id_Y$ and $p \circ q \cong id_M$ via the homotopy map
$$H_t: M \times I \to M, (x, s,t) \mapsto (x, t +s(1-t)), y \mapsto y$$
Then: $H_0= id_M, H_1= p \circ q$