From the isomorphism \(\pi_1(X \times Y, (x_0, y_0)) \approx \pi_1(X, x_0) \times \pi_1(Y, y_0)\) 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.
Solution
Let \[ a:I\to X, \qquad b:I\to Y \] be loops based at \(x_0\) and \(y_0\), respectively. Write \[ \alpha(t)=(a(t),y_0), \qquad \beta(t)=(x_0,b(t)). \]
1. Define\[ G:I^2\to X\times Y, \qquad G(u,v)=(a(u),b(v)). \]
The map is continuous as the product of the two continuous coordinate maps. Along the bottom and top edges it traces \(\alpha\), while along the left and right edges it traces \(\beta\), since \(a(0)=a(1)=x_0\) and \(b(0)=b(1)=y_0\).
2. Parameterize the bottom-then-right and left-then-top paths in the square by\[ p_0(s)= \begin{cases} (2s,0),&0\le s\le\frac12,\\ (1,2s-1),&\frac12\le s\le1, \end{cases} \]and\[ p_1(s)= \begin{cases} (0,2s),&0\le s\le\frac12,\\ (2s-1,1),&\frac12\le s\le1. \end{cases} \]
Both are continuous paths from \((0,0)\) to \((1,1)\). Under \(G\), the first becomes the concatenation \(\alpha\cdot\beta\), while the second becomes \(\beta\cdot\alpha\).
3. For \(t\in I\), set\[ p_t(s)=(1-t)p_0(s)+t p_1(s). \]Then\[ H(s,t)=G(p_t(s)) \]is a homotopy relative to endpoints from \(\alpha\cdot\beta\) to \(\beta\cdot\alpha\).
The square \(I^2\) is convex, so \(p_t(s)\in I^2\) for all \((s,t)\). The map \((s,t)\mapsto p_t(s)\) is continuous, hence so is \(H\). At \(t=0\) and \(t=1\) one has \[ H(s,0)=G(p_0(s))=(\alpha\cdot\beta)(s), \] and \[ H(s,1)=G(p_1(s))=(\beta\cdot\alpha)(s). \] Moreover, \[ p_t(0)=(0,0), \qquad p_t(1)=(1,1) \] for every \(t\), and both points map under \(G\) to \((x_0,y_0)\). Thus the endpoints remain fixed throughout the homotopy.
4. Therefore the classes of \(\alpha\) and \(\beta\) commute in \(\pi_1(X\times Y,(x_0,y_0))\).
The based homotopy in 3 gives \[ [\alpha][\beta]=[\alpha\cdot\beta]=[\beta\cdot\alpha]=[\beta][\alpha]. \]