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
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 $1$.
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 $\sum_i b_i$.