Commuting elements in product fundamental group

Question

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.

Answer 1

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.

Answer 2

I see now what my problem was, I never understood the complete formulas for the constant loop homotopy, but I found it in Lee. We have a homotopy $a \sim x_0 \cdot a$ by

$H_t(s) = \begin{cases} x_0 & 0 \leq s \leq t/2 \\ a(\frac{2s - t}{2 - t}) & t/2 \leq s \leq 1 \end{cases}$

and for the other side we have a homotopy $b \sim b \cdot y_0$ given by

$G_t(s) = \begin{cases} b(\frac{2s}{2 - t}) & 0 \leq s \leq 1 - t/2 \\ y_0 & 1 - t/2 \leq s \leq 1 \end{cases}$.

Thus the homotopy we are looking for then is $J_t(s) = (H_t(s), G_t(s))$. $J_0(s) = (a(s),b(s))$, $J_1(s) = ((x_0 \cdot a)(s), (b \cdot y_0)(s))$ and $J_t(0) = J_t(1) = (x_0, y_0)$. We can construct a similar based homotopy from $(a \cdot x_0, y_0 \cdot b)$ to $(a, b)$ and since homotopy is an equivalence relation we have finally have a homotopy from $(a \cdot x_0, y_0 \cdot b)$ to $(x_0 \cdot a, b \cdot y_0)$.