Here is the question I am trying to understand its solution:
From the isomorphism $\pi_1(X \times Y, (x_0, y_0)) \cong \pi_1(X, x_0) \times \pi_1(Y, y_0)$ it follows that loops in $X \times \{y_0\}$ and $x_0 \times \{Y\}$ represent commuting elements of $\pi_1(X \times Y, (x_0, y_0)).$ Construct an explicit homotopy demonstrating this.
And here is one of the solutions given in the following link (I am looking for the second solution: ) Commuting elements in product fundamental group:
"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 $(x_0 \cdot a, b \cdot y_0)$."
My question is:
Why in the homotopy between $b$ and $b.y_0,$ which is $G_s(t),$ why in the first part $s \leq 1 – \frac{t}{2},$ could someone explain this to me please?
Best Answer
One "reason" is that: $$\frac{2s}{2-t}>\frac{2-t}{2-t}=1$$When $s>1-t/2$ so $b(2s/2-t)$ isn't defined. $1-t/2$ is the point at which we stop, giving $b(1)=y_0$, in the user's homotopy.