How to prove/show this actually defines a homomoprhism 
We define the  homomorphism $f: \text{SL}_2(\mathbb Z / 2 \mathbb Z) \to \text{SL}_2(\mathbb Z / 2 \mathbb Z)$  that maps the generators to:

$ \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
$ \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \to \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$
How do we know this is a homorphism? In a previous exercise I explored that these two elements generate the group, the first has order 3, the second has order 2. Essentially this maps any power of the first matrix to the identity, and any power of the second to a power of the matrix $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ is it as simple as:
$$f\left(\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}^{m \bmod 3}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}^{n \bmod 2}\right)=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}^{n \bmod 2}$$
But since the matrices don't commute I find it hard to prove this is a homomorphism ($f(AB)=f(A) f(B))$.

Essentially I want to prove that:
$f$ defined by:
$$\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix},\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}^2= \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}^3 =\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$
$$\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}    \to \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} $$
is a homomorphism.
 A: Note that$$\mathrm{SL}_2(\mathbb{Z}_2)=\left\{\begin{bmatrix}1&0\\0&1\end{bmatrix},\begin{bmatrix}1&1\\1&0\end{bmatrix},\begin{bmatrix}1&0\\1&1\end{bmatrix},\begin{bmatrix}1&1\\0&1\end{bmatrix},\begin{bmatrix}0&1\\1&1\end{bmatrix},\begin{bmatrix}0&1\\1&0\end{bmatrix}\right\}.$$Here, the first element is $\operatorname{id}_{\mathrm{SL}_2(\mathbb{Z}_2)}$, the next three have order $2$ and the last two ones have order $3$. So, $\mathrm{SL}_2(\mathbb{Z}_2)\simeq S_3$. And your question basically becomes: is there an endomorphism of $S_3$ whose kernel contains an element of order $3$ and such that an element $g$ of order $2$ is mapped into itself? Yes, there is. First you consider the projection $S_3\longrightarrow S_3/A_3\simeq S_2$ and then you composite it with the homomorphism from $S_2$ into $S_3$ which maps the non-identity element into $g$. Then you will get the map\begin{align}\operatorname{id}&\mapsto\operatorname{id}\\(1\ \ 2)&\mapsto(1\ \ 2)\\(1\ \ 3)&\mapsto(1\ \ 2)\\(2\ \ 3)&\mapsto(1\ \ 2)\\(1\ \ 2\ \ 3)&\mapsto\operatorname{id}\\(1\ \ 3\ \ 2)&\mapsto\operatorname{id}.\end{align}You can now translate this back into $\textrm{SL}_2(\mathbb{Z}_2)$.
A: A very concrete way to verify that this is a homomorphism, despite the fact that the generators don't commute, is to find another relation between the generators. This approach turns out to be effective quite often. In this case, setting
$$T:=\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}
\qquad\text{ and }\qquad
S:=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},$$
a quick computation shows that $ST=T^2S$. You already know that $T$ and $S$ have order $3$ and $2$, respectively, so this shows that $ST=T^{-1}S$, so in general for integers $a$ and $b$ this yields
$$S^bT^a=T^{(-1)^ba}S^b.$$
It also shows that for every $M\in\operatorname{SL}_2(\Bbb{Z}/2\Bbb{Z})$ there exists unique $a\in\{0,1,2\}$ and $b\in\{1,2\}$ such that $M=T^aS^b$, and moreover that
$$(T^cS^d)(T^aS^b)=T^c(S^dT^a)S^b=T^c(T^{(-1)^da}S^d)S^b=T^{(-1)^da+c}S^{b+d}.$$
Now it is easy to check that $f$ is a homomorphism, because
$$f((T^cS^d)(T^aS^b))=f(T^{(-1)^da+c}S^{b+d})=I^{(-1)^da+c}S^{b+d}=S^{b+d}$$
$$f(T^cS^d)f(T^aS^b)=(I^cS^d)(I^aS^b)=S^dS^b=S^{b+d}.$$
