Prove that $\mathbb{Q}$ is not projective using the Hom functor I am aware of more direct proofs of this on this website but I am looking for a proof along the following lines:
Find an epimorphism $f:A \to B$ such that $Hom(\mathbb{Q},A) \to Hom(\mathbb{Q},B)$ is not an epimorphism. Therefore the above Hom functor is not exact and $\mathbb{Q}$ is not projective. I am thinking that I should use the fact that $Hom(\mathbb{Q},\mathbb{Z})=0$.
 A: This is continuing from Lord Shark's answer. You might not be satisfied with this proof because you have to choose... Nonetheless...,
Proof: Choose a generating set $S$ for $\mathbb Q$ as a $\mathbb Z$-module (for example $S = \mathbb Q$ works, but there are (I think) smaller sets you can choose). Take $A$ to be the free abelian group with $\mathbb Z$-basis $S$, so $A \cong \bigoplus_{s \in S} \mathbb Z$, and take $B = \mathbb Q$; since $S$ generates $\mathbb Q$, there is an surjective $\mathbb Z$-module homomorphism $A \to \mathbb Q$. Now $$Hom(\mathbb Q, A) \cong Hom(\mathbb Q, \bigoplus_{s\in S}\mathbb Z) \cong \bigoplus_{s \in S}Hom(\mathbb Q, \mathbb Z) = \bigoplus_{s \in S} 0 = 0$$ where the second isomorphism is from a basic fact that $Hom(-, -)$ commutes with direct sums in the second slot; and the second to last equality is by the fact OP stated.
On the other hand $Hom(\mathbb Q, \mathbb Q) \neq 0$ since it contains the identity map for example. $\square$
A: You can take $A$ to be a free Abelian group and $B$ to be $\Bbb Q$.
