The fact that $\mathbb Z/N \to\mathbb Z/n \times \mathbb Z/m$ where $N=m \cdot n$, and both coprime is known as the chinese remainder theorem.
The fact that it is injective, is an application of the first isomorphism theorem for groups, since the map $\varphi:\mathbb Z \to \mathbb Z/m \times \mathbb Z/n$ given by $a \mapsto(a,a)$ is in the kernel if and only if $m \mid a$ and $n \mid a$ if and only if $N \mid a$, or in other words, $a \in N \mathbb Z$, so $\mathbb Z/\ker \varphi=\mathbb Z/N$.
In your case, you can get your hands dirty:
hint for surjectivity: for surjectivity, show that there exists some $x \in \mathbb Z/6$ so that $x \equiv a_1 \mod 2$ and $x \equiv a_2 \mod 3 $ for any choice of $(a_1,a_2) \in \mathbb Z/2 \times \mathbb Z/3$. One way to approach this, is to find integers $m,n$ so that $2m+3n=1$ and set $x=3a_1n+2a_2m$ and reduce mod $6$.
Hint for injectivity: reproduce the argument given in the second paragraph, but specialize it for your problem.
Hint if you're feeling lazy: take whichever problem (injectivity/ surjectivity) you find easier, and note that the two groups have the same cardinality, so either condition implies the other.