# Commuting elements in product fundamental group

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.

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.
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 $$(x_0 \cdot a, b \cdot y_0)$$.