What is the kernel of this map $\Phi: F_2 \to \mathbb{Z}_2 \oplus \mathbb{Z}_3$? What is the kernel $K\leq F_2 = \langle a,b \rangle$ of this map $\Phi: F_2 \to \mathbb{Z}_2 \oplus \mathbb{Z}_3$ given by $a \mapsto (1+2\mathbb{Z},0+3\mathbb{Z})$ and $b\mapsto (0+2\mathbb{Z}, 1+3\mathbb{Z})$?
Here is my thought process so far:
Since $K$ is the kernel, it will be all words on $F_2$ such that their image under $\Phi$ is the identity $e$, which in this case would be $(0+2\mathbb{Z}, 0+3\mathbb{Z})$.
I see that clearly $a+a =e$, as well as $4a$ or $n\cdot a$ for $n$ a multiple of 2.
Similarly, $b+b+b=e$, and so on for multiples of 3.
Any combination of these would also obtain $e$, so my initial thought is that $K$ is generated by $a^2$ and $b^3$.
For instance, $\Phi(a^4b^9)=e+e=e$.
Would this be the correct way of stating it?
 A: Your logic is almost correct, though you should justify why you found all of the elements of the kernel of $\Phi$. You have only justified the reason why $\langle a^2,b^3\rangle$  belongs in the kernel, but not why there is nothing else in it. In fact, it is not true that these are the only elements in the kernel: for example $ab^2ab$ is in the kernel too!
You can argue as follows: any element of $F_2$ is by definition a word in terms of $a$ and $b$. Because $\mathbb{Z}_2\oplus\mathbb{Z}_3$ is an abelian group, the image of $a$ and $b$ under $\Phi$ commute in $\mathbb{Z}_2\oplus\mathbb{Z_3}$. Thus, any $x\in F_2$ has image of the form $$\Phi(x) = m\Phi(a)+n\Phi(b) = (m+2\mathbb{Z},n+3\mathbb{Z})$$
for integers $m,n$ which correspond to the total power of $a,b$ respectively that appears in the word $x$. If $x$ is in the kernel, then this $\Phi(x)$ must of course be zero, which means that $m$ is even and $n$ is divisible by $3$, so we are done.
(Also, you write $a+a=e$ which is not correct; it should rather be $\Phi(a)+\Phi(a)=e$ where we are working in the group $\mathbb Z_2\oplus\mathbb Z_3$. However it is also conventional to denote the identity of an additive group by $0$ instead of $e$.)
A: I assume that by $F_n$ you mean free group, not free abelian group. It would be unusual to denote a free abelian group by $F_n$.

Any combination of these would also obtain $e$, so my initial thought is that $K$ is generated by $a^2$ and $b^3$.

That is not enough. First an example. Consider the general linear group of square matrices over $\mathbb{R}$: $GL(2,\mathbb{R})$. Now pick two matrices
$$A=\left[\begin{matrix}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1 \\
\end{matrix}\right]$$
$$B=\left[\begin{matrix}
1 & 0 & 0 \\
0 & \cos\big(\frac{2}{3}\pi\big) & -\sin\big(\frac{2}{3}\pi\big) \\
0 & \sin\big(\frac{2}{3}\pi\big) & \cos\big(\frac{2}{3}\pi\big) \\
\end{matrix}\right]$$
Both are rotations in $\mathbb{R}^3$, but around different axis. Furthermore you can verify that $A^2=1$ and $B^3=1$ but $AB\neq BA$, and so the subgroup generated by $\langle A,B\rangle$ cannot be isomorphic to $\mathbb{Z}_2\oplus\mathbb{Z}_3$ even though it does satisfy $A^2=1$ and $B^3=1$ relations.
This shows that $\langle a,b\ |\ a^2=1,\ b^3=1\rangle$ is not a presentation of $\mathbb{Z}_2\oplus\mathbb{Z}_3$.

But you are very close. In fact, it is enough to add commutativity relations. More precisely your $K$ is generated by $\langle a^2, b^3, aba^{-1}b^{-1}\rangle$. How do we know that? Well, consider the following group homomorphism:
$$g:\langle a,b\ |\ a^2=1,\ b^3=1,\ ab=ba\rangle\to \mathbb{Z}_2\oplus\mathbb{Z}_3$$
$$g(a)=(1,0)$$
$$g(b)=(0,1)$$
I leave as an exercise that if my $g$ is an isomorphsim then your $K=\langle a^2,b^2,aba^{-1}b^{-1}\rangle$.
It clearly is surjective. Now assume that $g(x)=0$. Since our group on the left is abelian, then $x=a^nb^m$. It follows that $0=g(x)=g(a)^ng(b)^m=(n,m)$ which is if and only if $2|n$ and $3|m$. But then $a^n=1$ and $b^m=1$, meaning $x=1$ and thus $\ker g$ is trivial. Which shows that $g$ is an isomorphism.
