2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.