- \(\Rightarrow\) Let \([\alpha] \in \pi_1(X\cross Y, (x_0, y_0))\) be an arbitrary loop in \(X \cross Y\). Then \(\alpha\) is equivalently a map \(S^1 \into X \cross Y\). Considering \(S^1\) to be a subset of \(\RR^2\), we can parameterize \(\alpha\) as \(\alpha(z) = \alpha(x+iy) = (\alpha_x(x), \alpha_y(y))\) in components. In particular, since \(\alpha\) is continuous, so are \(\alpha_x, \alpha_y\). Moreover, since \(\alpha(0) = \alpha(0 + i0) = (x_0, y_0)\), we have \(\alpha_x(0) = x_0, \alpha_y(0) = y_0\). (Note: alternatively, given the product, we have projections \(p_X, p_y\), so we can define the map \(\alpha \mapsto (p_X \circ \alpha, p_Y \circ \alpha)\))
But then \(\alpha_x: S^1 \into X\) and \(\alpha_y: S^1 \into Y\) are loops entirely in \(X, Y\) at the respective base points, and so we can define the map \(F: \pi_1(X\cross Y, (x_0, y_0)) \into \pi_1(X, x_0) \cross \pi_1(Y, y_0)\) by \([\alpha] = [(\alpha_x, \alpha_y)] \mapsto ([\alpha_x], [\alpha_y])\)
This is injective, since \(([a],[b]) = ([c],[d])\) on the RHS means that \([a] = [c], [b] = [d]\) in the fundamental groups, and thus \(a\homotopic c, b\homotopic d\) in the spaces. We want to show that \([(a,b)] = [(c,d)]\), which would follow if \(\alpha(x+iy) = (a(x),b(y)) \homotopic \beta(x+iy) = (c(x),d(y))\) in \(X\ \cross Y\). …?
This is surjective, because if \(([a], [b])\) are elements in the right-hand side, then \(a(0) = a(1) = x_0\) and \(b(0) = b(1) = y_0\), so we can consider \((a,b): I \into X \cross Y\) where \((a,b)(z) = (a,b)(x+iy) = (a(x), b(y))\). This is then a loop in \(X\cross Y\), since \((a,b)(0) = (a(0), b(0)) = (0,0) = (x_0, y_0)\) and similarly \((a,b)(1) = (a(1), b(1)) = (x_0, y_0)\). So this is actually a map \((a,b): S^1 \into X \cross Y\), or in other words, a loop in \(X\times Y\) based at \((x_0, y_0)\), which lifts to an element of the fundamental group on the LHS.
Maps in both directions are continuous, since a vector function is continuous iff its component functions are continuous.
This is well-defined, due to the fact that if \(a \homotopic b\), then \(p_X \circ a \homotopic p_X\circ b\), and \(F = (f_x, f_y)\) is a homotopy iff its components functions are homotopies.
Solution
1. Define\[\Phi:\pi_1(X\times Y,(x_0,y_0))\to\pi_1(X,x_0)\times\pi_1(Y,y_0),\quad [\alpha]\mapsto([p_X\alpha],[p_Y\alpha]).\]
Projection preserves based homotopies, so the map is well-defined; it is a homomorphism because projections commute with concatenation.
2. Define the inverse by\[\Psi([a],[b])=[t\mapsto(a(t),b(t))].\]
Componentwise based homotopies combine to a based homotopy in the product, so this is well-defined and multiplicative.
3. The maps \(\Phi\) and \(\Psi\) are inverse isomorphisms.
A map into a product is exactly the ordered pair of its coordinate maps, so both composites are identities.
4. Hence\[\boxed{\pi_1(X\times Y,(x_0,y_0))\cong\pi_1(X,x_0)\times\pi_1(Y,y_0).}\]
This is <1>1–<1>3.