Working on Homotopy Equivalent Suppose I have spaces $X_{0}, X_{1},Y_{0},Y_{1}$ and $X_{0}$ is homotopy equivalent to $Y_{0}$ and $X_{1}$ is homotopy equivalent to $Y_{1}$.
I would like to show that $X_{0}$ x $X_{1} $ is homotopy equivalent to $Y_{0}$ x $Y_{1} $.
How do I approach this statement? I am fresh new to this topic so any input will be greatly appreciated.
Thanks!
 A: Since $X_0 \simeq Y_0$ and $X_1 \simeq Y_1$, there are two pairs of functions
$$X_0 \overset{f_0}{\underset{g_0}{\rightleftarrows}} Y_0 \quad \text{and} \quad X_1 \overset{f_1}{\underset{g_1}{\rightleftarrows}} Y_1$$
such that the relevant composites are homotopic to the relevant identities.
From these functions, you can construct functions
$$X_0 \times X_1 \overset{f}{\underset{g}{\rightleftarrows}} Y_0 \times Y_1$$
where $f$ and $g$ are defined by
$$f(x_0,x_1) = (f_0(x_0),f_1(x_1)) \quad \text{and} \quad g(y_0,y_1)=(g_0(y_0),g_1(y_1))$$
for all $x_0,x_1,y_0,y_1$.
You need to check that $g \circ f \sim \mathrm{id}_{X_0 \times X_1}$ and $f \circ g \sim \mathrm{id}_{Y_0 \times Y_1}$.
In a similar way to how we constructed $f$ and $g$ from $f_0,f_1,g_0,g_1$, you can construct the two desired homotopies from the homotopies
$$g_0 \circ f_0 \sim \mathrm{id}_{X_0}, \quad g_1 \circ f_1 \sim \mathrm{id}_{X_1}, \quad f_0 \circ g_0 \sim \mathrm{id}_{Y_0} \quad \text{and} \quad f_1 \circ g_1 \sim \mathrm{id}_{Y_1}$$
I'll leave you to construct these homotopies.
A: acord your homotopy you to functions f0 and f1 and g0 g1 and g0(f0) in X0 and(g1(f1)) in X1 are nullhomotopys. I try to take the function f:=(f0,f1) and g:=(g0,g1) so g(f)=(g0(f0),g1(f1)) is nullhomotopy in X0*X1.
