From Hatcher 1.1.10:
From the isomorphism $\pi_1(X \times Y, (x_0, y_0)) \simeq \pi_1(X,x_0) \times \pi_1(Y,y_0)$ it follows that loops in $X \times \lbrace x_0 \rbrace$ and $\lbrace x_0 \rbrace \times Y$ represent commuting elements of $\pi_1(X \times Y, (x_0, y_0))$. Construct an explicit homotopy demonstrating this.
My thinking thus far is that if $a(s)$ is a loop in $X \times \lbrace y_0 \rbrace$ then it induces a loop in $X \times Y$, $(a(s), y_0)$, and similarly for a loop $b(s)$ in $Y \times \lbrace x_0 \rbrace$. It is clear that I want to show that $[(a(s), y_0)] \cdot [(x_0, b(s))] = [(a(s), b(s))] = [(x_0, b(s))] \cdot [(a(s), y_0)]$ but I'm at a loss on how to create a homotopy showing this. Any hints would be appreciated.
Best Answer
Let $e$ denote the constant loop. We have $(a,b)\sim (a * e , e * b)= (a,e) * (e,b)$ and similarly $(a,b)\sim (e * a , b * e)= (e,b) * (a,e)$. So to construct an explicit homotopy you just compose the homotopies $(a * e, e*b) \sim (a,b)$ and $(a,b)\sim (e * a , b * e)$ which are just products of the normal homotopies involving loop composition with the constant loop.