Theorem: Let $i^x: X \rightarrow X \times Y$ and $i^y: Y \rightarrow X \times Y$ be the maps $x \mapsto (x, y_0)$ and $y \mapsto (x_0, y)$ for some fixed $x_0, y_0$. The mapping $\varphi: \pi(X,x_0) \times \pi(Y,y_0) \rightarrow \pi(X \times Y, (x_0, y_0))$ by $\varphi: (\beta, \gamma) \mapsto (i^x_* \beta) \cdot (i^y_* \gamma)$ is an isomorphism.

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Proof: Consider the map defined in Theorem 7.1 -- that is, $\phi : \pi(X \times Y, (x_0, y_0)) \rightarrow \pi(X, x_0) \times \pi(Y, y_0)$ by $\alpha \mapsto (p^x_* \alpha, p^y_* \alpha)$, where $p^x$ and $p^y$ are the projection maps. This map is an isomorphism. We will show that $\phi \varphi$ is the identity, so that $\varphi = \phi^{-1}$. Let $\beta \in \pi(X,x_0)$ and $\gamma \in \pi(Y,y_0)$. Then,

$$\phi(\varphi(\beta, \gamma)) = \phi(i^x_* \beta \cdot i^y_* \gamma) = \Big( p^x_* \big( i^x_* \beta \cdot i^y_* \gamma \big), p^y_* \big( i^x_* \beta \cdot i^y_* \gamma \big) \Big)$$

Now, note that $(i^x_* \beta)(t) = (\beta(t), y_0)$, and similarly for $i^y_* \gamma$, so that the product path is:

$$i^x_* \beta \cdot i^y_* \gamma = \left\{ \begin{array}{lr} (\beta(2t), y_0) & 0 \leq t \leq 1/2 \\ (x_0, \gamma(2t-1)) & 1/2 \leq t \leq 1 \end{array} \right.$$

If we apply $p^x_*$, the second half of the path will effectively be nullified -- that is:

$$p^x_* \big( i^x_* \beta \cdot i^y_* \gamma \big) = \left\{ \begin{array}{lr} \beta(2t) & 0 \leq t \leq 1/2 \\ x_0 & 1/2 \leq t \leq 1 \end{array} \right.$$

Clearly, this is a path in $\pi(X,x_0)$ that traverses $\beta$ twice as fast, and as such, it's clearly homotopic to $\beta$. An explicit homotopy is:

$$F(t,s) = \left\{ \begin{array}{lr} \beta((2-s)t) & 0 \leq t \leq \frac{1}{2} (1+s) \\ x_0 & \frac{1}{2} (1+s) \leq t \leq 1 \end{array} \right.$$

A very similar argument proceeds for the other path in the ordered pair, with the first half of the path being nullified and using the 'reverse' homotopy. We then obtain

$$p^x_* \big( i^x_* \beta \cdot i^y_* \gamma \big) \sim \beta \qquad p^y_* \big( i^x_* \beta \cdot i^y_* \gamma \big) \sim \gamma$$

So, the components of $(\phi \varphi)(\beta, \gamma)$ are in the equivalence class of $\beta$ and $\gamma$, respectively, so $\phi \varphi$ is the identity on $\pi(X, x_0) \times \pi(Y, y_0)$. Then, it must be that $\varphi = \phi^{-1}$, and is itself an isomorphism.

$$\mjqed$$

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It follows that the elements $i^x_* \beta$ and $i^y_* \gamma$ commute. It's easy to see that $\phi(i^x_* \beta \cdot i^y_* \gamma) = (\beta, \gamma)$, and also $\phi(i^y_* \gamma \cdot i^x_* \beta) = (\beta, \gamma)$. So it must be that $i^x_* \beta \cdot i^y_* \gamma = i^y_* \gamma \cdot i^x_* \beta$.