Show that the fundamental group of wedge of two circles is non Abelian without actually computing it.

Question

Given that there is a pointed space $(Y,y_0)$, such that it has non Abelian fundamental group, need to show that the fundamental group of the wedge of two circles is non Abelian.

What I was thinking was that if I could define a surjective homomorphism from $\pi_1(S^1 V S^1)$ to $\pi_1(Y)$, then I'd be done.

To that end, I thought it'd be more intuitive to think of loops at the basepoint as continuous maps from $S^1$ rather than as paths from $[0,1]$, but I'm still having trouble coming up with the actual homomorphism, since the space $Y$ is arbitrary.

Answer 1

You don't need a surjective homomorphism from $\pi_1(S^1\vee S^1,p)$ to $\pi_1(Y,y_0)$. You just need a homomorphism whose image contains two noncommuting elements of $\pi_1(Y)$. With that in mind, let $[a],[b]\in\pi_1(Y,y_0)$ be two noncommuting loops. How might you use $a$ and $b$ to define a map $S^1\vee S^1\to Y$?

More details are hidden below.

Just take the map $f:S^1\vee S^1\to Y$ which is given by the loop $a$ on the first copy of $S^1$ and is given by the loop $b$ on the second copy. Then $f$ gives a homomorphism $f_*:\pi_1(S^1\vee S^1,p)\to\pi_1(Y,y_0)$, where $p$ is the wedge point. Also, $[a]$ and $[b]$ are in the image of $f_*$, since they are just $f_*([i])$ and $f_*([j])$ where $i$ and $j$ are the loops in $S^1\vee S^1$ that go around each of the circles.