Show that if $Y$ is contractible, then the canonical projection $\pi: X\times Y\to X$ is a homotopy equivalence

Question

Show that if $Y$ is contractible, then the canonical projection $\pi:X\times Y\to X$ is a homotopy equivalence.

Answer 1

To prove that $\pi$ is a homotophy equivalence we need to find some continuous map $\tau$ such that $$\tau \circ \pi \simeq \operatorname{id}_{X\times Y}\\ \pi \circ \tau \simeq \operatorname{id}_{X}$$ But we also know that since $Y$ is contractible, $Y$ is homotophy equivalent to a point, so we have continuous maps $r : Y \to \{0\}$ and its inverse $r^{-1}:\{0\} \to Y$, which is a homotopy equivalence. So we consider the composition $$X\times Y \xrightarrow{(\operatorname{id}_{X},r)}X\times \{0\}\xrightarrow{\pi'}X$$ Where $(\operatorname{id}_{X},r):X\times Y \to X\times \{0\}$ sends $(x,y)$ to $(x,r(y))$. We can show that this is a homotopy equivalence with inverse $(\operatorname{id}_{X},r^{-1})$. These make a homotopy equivalence because $$(\operatorname{id}_{X},r^{-1})\circ (\operatorname{id}_{X},r)\simeq(\operatorname{id}_{X},\operatorname{id}_{Y})=\operatorname{id}_{X\times Y}\\ (\operatorname{id}_{X},r)\circ (\operatorname{id}_{X},r^{-1})\simeq(\operatorname{id}_{X},\operatorname{id}_{\{0\}})=\operatorname{id}_{X\times \{0\}}$$ We can also define $\pi':(x,0)\mapsto x$ and its inververse $\pi'^{-1}:x\mapsto (x,0)$ which is a homeomorhpism and therefore also a homotophy equivalence.

We also have that $\pi = \pi'\circ (\operatorname{id}_{X},r)$, and the composition of homotopy equivalencies is also a homotophy equivalence. Explicitly, we can construct $\tau$ as $$(\operatorname{id}_{X},r^{-1})\circ \pi'^{-1}$$

And it's easy to see that our conditions for homotopy equivalence follow.