I'm attempting to show that the $n$-fold torus (i.e. the connected sum of $n$ tori, $T_1 \# T_2 \# \ldots \# T_n$) has non-abelian fundamental group when $n > 1$.
It has been suggested that I consider mapping the free group on the $2n$ letters $\alpha_1, \beta_1, \ldots, \alpha_n, \beta_n$ (where $\alpha_i, \beta_i$ are generators of $\pi_1(T_i, b)$) to the free group on $2$ letters $\gamma, \delta$, such that $\alpha_1, \beta_1 \to \gamma$ and $\alpha_i, \beta_i \to \delta$ for $i \neq 1$.
I am able to show that this map is a homomorphism. And I can show that if there were a homomorphism from $\pi_1(T\#\ldots\# T, b)$ to a non-abelian group (such as the free group on $\gamma, \delta$), then $\pi_1(T\#\ldots\# T, b)$ would necessarily be non-abelian itself. But $\pi_1(T\#\ldots\# T, b)$ is not the free group on 2n letters, but rather this group mod the least normal subgroup containing $[\alpha_1, \beta_1] \cdots [\alpha_n, \beta_n]$.
I've also attempted another approach, via considering the $n$-fold torus as a $4n$-gon with appropriate identifications of edges, but wasn't able to draw anything concrete from that. Perhaps I simply lacked the appropriate geometric intuition there?
Best Answer
As you point out, $\pi_1(\Sigma_g) \cong \langle \alpha_1, \beta_1, \dots, \alpha_g, \beta_g \mid [\alpha_1, \beta_1] \cdots [\alpha_g, \beta_g]\rangle$. Define a homomorphism from this to the free group on two letters by $\alpha_1 \mapsto x_1$, $\beta_1 \mapsto x_2$, $\alpha_2 \mapsto x_2, \beta_2 \mapsto x_1$, and all other generators map to $1$. Because $[x_1,x_2][x_2,x_1] = 1$, this does indeed define a homomorphism $\pi_1(\Sigma_g) \to F_2$, and because each of the generators are mapped to, it's surjective, so $\pi_1(\Sigma_g)$ is nonabelian.
You can see explicitly why this fails for $g=1$.