§ Covariant Hom is left exact

Let's say we have the exact sequence:
$0 \rightarrow A \xrightarrow{i} B \xrightarrow{\pi} C \rightarrow 0$
Where the first arrow is $i$ for inclusion and the third is $\pi$ for projection. We now want to consider what happens when we have $Hom(X, -)$ for some target space $X$. The induced arrows are induced from composition; I will write $(f; g)(x) \equiv g(f(x))$ to mean "first do $f$, then do $g$". Hence, my arrows linking $Hom(X, A)$ to $Hom(X, B)$ will be $-;i$: To first go from $X$ to $A$, and then apply $i$ to go from $A$ to $B$. This gives us the sequence (which we are to check if it is exact, and which of the left and right arrow exist):
$0 \xrightarrow{?} Hom(X, A) \xrightarrow{-;i} Hom(X, B) \xrightarrow{;-\pi} Hom(X, C) \xrightarrow{?} 0$

§ A particular example

As usual, we go to the classic exact sequence:
$0 \xrightarrow 2Z \rightarrow{\pi} Z \rightarrow{\pi} Z/2Z \rightarrow 0$
We now have three interesting choices for our $X$ in relation to the above sequence: (a) $Z$, (b) $2Z$, (c) $Z/2Z$. Since $Z$ and $2Z$ are isomorphic as modules, let's study the case with $Z$ and $Z/2Z$