Let $X$ be a path-connected topological space. And there is a continuous map $F: X\times X \to X$ such that: $$F(x,x)=x \ \text{ and }F(x,y)=F(y,x).$$ Prove: The fundamental group of $X$ is Abelian.
This problem is in my topology class homework. We can show that if f,g are loops in X, then F(f,g) is also a loop in X. Let [f] be the element in the fundamental group represented by f, and e be the constant map which represents the unit element in the fundamental group. Then for any [f] and [g], we have [F(f,e)][F(g,e)]=[F(g,e)][F(f,e)]. It turns out that [f]->[F(e,f)] is a group homomorphism. And the image of this homomorphism is an Abelian group. However, I cannot show that this homomorphism is surjective, which is needed to solve this problem.
Guys I've come up with a solution. Use [F(f,e)][F(g,e)]=[F(g,e)][F(f,e)] and x=F(x,x)=F(e,x)F(e,x).
To see that $F_*$ is surjective, observe that since any curve $f$ in $X$ lifts to the curve $(f,f)$ in $X\times X$, any loop $[f]\in \pi_1(X)$ lifts to $([f],[f])\in \pi_1(X)\times\pi_1(X)\cong\pi_1(X\times X)$. Since $F(x,x) = x$, you know that $F_*([f],[f]) = [f]$.