Hom(G,-) functor is exact? Given an exact sequence $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ we know that the induced sequence $0 \rightarrow Hom(G,A)\rightarrow Hom(G,B)\rightarrow Hom(G,C)$ is exact. But wouldn't the last homomorphism be onto since functors preserve injectivivity and surjectivity? And thus you could extend the sequence by $Hom(G,C) \rightarrow 0$ and preserve exactness? It would follow that $Hom(G,-)$ is exact. What is my mistake?
 A: The mistake is when you say that functors preserve injectivity and surjectivity. That is certainly not true for general functors, and even representable covariant functors $X\mapsto Hom(A,X)$ only preserve monomorphisms in general. In fact such a functor preserves epimorphisms if and only if the representing object $A$ is projective (by definition).
For example, in the category of abelian groups, consider the surjective map $ \mathbb{Z}\to \mathbb{Z}/4\mathbb{Z}$
and the functor represented by $\mathbb{Z}/2\mathbb{Z}$. Giving a map $\mathbb{Z}/2\mathbb{Z}\to A$ is the same thing as choosing an element of $A$ whose order divides $2$, so that $Hom(\mathbb{Z}/2\mathbb{Z} ,\mathbb{Z}/4\mathbb{Z}) \cong \mathbb{Z}/2\mathbb{Z}$, but $Hom(\mathbb{Z}/2\mathbb{Z},\mathbb{Z})\cong\{0\}$, thus the induced map between the two cannot be surjective.
A: I don't know in what category you are working but there are many where this property does not hold. 
In an abelian category, the objects $G$ for which the exactness of $\mbox{Hom}(G,-)$ hold are called projective objects. 
See this discussion here.
