[Math] The fundamental group of a product is the product of the fundamental groups of the factors

algebraic-topologyfundamental-groups

Hello 🙂 i want to prove the following statement:

  • $\pi_1(X\times Y,(x_0,y_0))\equiv\pi_1(X,x_0)\times\pi_1(Y,y_0)$

But how to do that? Is this just the projection and the use of the product topology?
Thank you for help 🙂

I also want to prove that the fundamental group of a n-sphere is trivial for $n>1$, but i have no idea. From my point of view homotopy is very difficult…

Best Answer

A loop $\alpha$ in $X\times Y$ is a continuous map $\ \alpha:S^1\to X\times Y$.

If $\ s\in S^1$ we can write $\alpha(s) = (\alpha_x(s),\alpha_y(s))$. By theorem, $\alpha$ is continuous if and only if both components $\alpha_x:S^1\to X$ and $\alpha_y:S^1\to Y$ are continuous (this applies for all maps $A\to X\times Y$. Here, $A=S^1$). Thus we have a bijection between loops $\alpha$ in $X\times Y$ and pairs of loops $(\alpha_x,\alpha_y)$ with $\alpha_x$ a loop in $X$ and $\alpha_y$ a loop in $Y$. If $\ p_x:X\times Y \to X$ and $p_y:X\times Y\to Y$ are the projections, this bijection is given by $\alpha \mapsto (p_x\circ \alpha,p_y\circ \alpha)$.

If $s_0\in S^1$, a map $\ \ f:S^1\times I\to X\times Y$, $\ \ \ (s,t)\mapsto (f_x(s,t),f_y(s,t))$ is a homotopy relative to $\{s_0\}$ if and only if both components $f_x$ and $f_y$ are homotopies rel $\{s_0\}$. Thus, loops $\alpha$ and $\beta$ in $X\times Y$ at $s_0$ are homotopic if and only if their projections to $X$ and to $Y$ are homotopic - $\alpha \simeq \beta$ iff $\ \ p_x\circ \alpha \simeq p_x\circ \beta$ and $p_y\circ\alpha \simeq p_y\circ\beta$.

Thus our bijection of loops induces a bijection $\pi_1(X\times Y) \to \pi_1(X)\times \pi_1(Y)$ given by the homomorphisms induced by the projections: $[\alpha]\mapsto ([p_x \circ \alpha],[p_y \circ \alpha])$. Since the maps induced by the projections are homomorphisms, the bijection is a homomorphism (simple fact about homomorphisms into product groups), and thus an isomorphism.

Related Question