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...
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$\begingroup$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.
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ 1 $\begingroup$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.
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