§ Covariant Hom is left exact

Let's say we have the exact sequence:
0AiBπC0 0 \rightarrow A \xrightarrow{i} B \xrightarrow{\pi} C \rightarrow 0
Where the first arrow is ii for inclusion and the third is π\pi for projection. We now want to consider what happens when we have Hom(X,)Hom(X, -) for some target space XX. The induced arrows are induced from composition; I will write (f;g)(x)g(f(x))(f; g)(x) \equiv g(f(x)) to mean "first do ff, then do gg". Hence, my arrows linking Hom(X,A)Hom(X, A) to Hom(X,B)Hom(X, B) will be ;i-;i: To first go from XX to AA, and then apply ii to go from AA to BB. This gives us the sequence (which we are to check if it is exact, and which of the left and right arrow exist):
0?Hom(X,A);iHom(X,B);πHom(X,C)?0 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:
02ZπZπZ/2Z0 0 \xrightarrow 2Z \rightarrow{\pi} Z \rightarrow{\pi} Z/2Z \rightarrow 0
We now have three interesting choices for our XX in relation to the above sequence: (a) ZZ, (b) 2Z2Z, (c) Z/2ZZ/2Z. Since ZZ and 2Z2Z are isomorphic as modules, let's study the case with ZZ and Z/2ZZ/2Z

§ Hom(;Z)Hom(-; Z)

§ Hom(;Z/2Z)Hom(-; Z/2Z)