# 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. 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. 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. 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.