# The fundamental group of a product is the product of the fundamental groups of the factors

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...

• I suggest making two questions out of that. – Julian Kuelshammer Jan 31 '13 at 11:52
• Your first question on the fundamental group of a product is dealt in chapter 1 of Hatcher (Proposition 1.12). Your second question can be proven using this question here: math.stackexchange.com/q/283532/38268 – user38268 Jan 31 '13 at 11:52
• Of course there is the more general, and useful, statement that the fundamental groupoid of the product of spaces is isomorphic to the product of the fundamental groupoids. – Ronnie Brown Jan 31 '13 at 12:06

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.

• "Thus we have a bijection $\pi_1(X\times Y)\rightarrow \pi_1(X) \times \pi_1(Y)$" doesn't follow from anything you've said previously. – PVAL-inactive Nov 13 '14 at 23:28

The proof of Owen for the first question is correct. Of course van Kampen works for the second question, but it is a slight overkill. I would prefer the following: $\pi_1(S^n)=0$ if and only if every loop based at $p$ is contractible through loops based at $p$. Given an arbitrary path based at $p$, we can easily find a contraction by stereographically projecting (from a point not lying on the path) it to $\mathbb R^n$, contracting it in $\mathbb R^n$ and composing the contraction with the inverse of the stereographic projection.

• This only works if the loop already misses a point, which is not the case in general. – Danu May 16 '17 at 10:05

Well, I guess it's more elegant to prove the second question using van Kampen theorem. Choose $U = S^n - P$ and $V= S^n - N$, where $P$ is the south pole and $N$ is the north pole. Since $n$ is bigger than 1, the intersection $U\cap V$ is connected. So, applying the van Kampen theorem, since $U$ and $V$ are contractible spaces, you get that $S^n$ has trivial fundamental group.