Surjective homomorphisms from Free group of rank 2 to Finite group of order 2/ I am currently revising for an exam and am unable to proceed on the following question. If anyone could help I would appreciate it.
I initially thought about trying to create a scenario using the fundamental groups of the Wedge sum of 2 circles and the projective plane, so I could apply SVK's theorem. Yet it seems like overkill.
Thankyou
 A: The question has actually nothing to do with algebraic topology. In fact it wouldn't be easy to use algebraic topology here, because while you know that the a continuous map $S^1 \vee S^1 \to \mathbb{RP}^2$ induces a morphisms on fundamental groups, you need to make sure that all morphisms can be obtained that way, and you need a way to know when two continuous maps induce the same map. Moreover you need to classify such continuous maps up to homotopy, and that's a bit of a circular argument.
Instead, there is a fully algebraic proof. Let $x,y \in F_2$ be generators of the free group $F_2$. Any morphism $f : F_2 \to G$ (where $G$ is an arbitrary group) is entirely determined by $f(x)$ and $f(y)$. Conversely, given $g,h \in G$, there exists a unique morphism $f : F_2 \to G$ such that $f(x) = g$ and $f(y) = h$.
So in your case you're looking at morphisms $f : F_2 \to \mathbb{Z}/2\mathbb{Z} = \{0,1\}$. You see that there are four cases depending on whether $f(x) = 0$ or $1$ and $f(y) = 0$ or $1$.


*

*If $f(x) = f(y) = 0$, then $f(X) = 0$ for all $X \in F_2$, hence $f$ is not surjective.

*Otherwise, either $f(x) = 1$ or $f(y) = 1$ or both, hence $f$ is surjective.


So you have three surjective morphisms $F_2 \to \mathbb{Z}/2\mathbb{Z}$. You can even write them down explicitly if you want. For example, the one characterized by $f(x) = f(y) = 1$ is given by the following map:
$$x^{a_1} y^{b_1} x^{a_2} y^{a_2} \dots x^{a_k} y^{b_k} \mapsto \sum_{i=1}^k a_i + b_i \pmod{2}$$
for $a_i, b_i \in \mathbb{Z}$. Two other two morphisms are respectively obtained by considering $\sum_i a_i$ and $\sum_i b_i$.
