Here is the problem(1.1.10 in Allen Hatcher Algebraic Topology) I want 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.
But before understanding its solution, I want to understand why 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 $\pi_1(X \times Y, (x_0, y_0)).$
Could someone explain this to me please? maybe by giving me a concrete example
Best Answer
Algebraically the isomorphism is saying a loop in $X\times\{y_0\}$, representing, say, the element $\alpha$, is identifiable with the element: $$(\alpha,1)\in\pi_1(X,x_0)\times\pi_1(Y,y_0)$$And similarly for $\{x_0\}\times Y$.
Such elements commute because: $$(\alpha,1)\cdot(1,\beta)=(\alpha,\beta)=(1,\beta)\cdot(\alpha,1)$$Passing back to $\pi_1(X\times Y)$, this will imply: $$\alpha\cdot\beta=\beta\cdot\alpha$$
The homotopy you need to construct can be thought of as mirroring this algebraic proof.