Projective Module equivalence proof

Question

Given an $A$-module $P$. How can I show that if there exists a module $Q$ such that $P \oplus Q$ is free then the functor $\operatorname{Hom}(P,\cdot)$ is exact?

Answer 1

This is a very standard argument.

Since $\DeclareMathOperator{\H}{Hom}\H(P,\cdot)$ is left exact, you only need to show it is right exact or, equivalently, that if $f\colon M\to N$ is surjective, then $f_*\colon\H(P,M)\to\H(P,N)$ is surjective.

Let $\alpha\in\H(P,N)$. Consider the canonical projection $p\colon P\oplus Q\to P$ and the homomorphism $\alpha\circ p\colon P\oplus Q\to N$. Let $B=\{x_i:i\in I\}$ be a basis of the free module $P\oplus Q$; for each $i\in I$ choose $m_i\in M$ such that $f(m_i)=\alpha\circ p(x_i)$, which is possible because $f$ is surjective.

Since $P\oplus Q$ is free with the basis $B$, there exists a unique homomorphism $\beta\colon P\oplus Q\to M$ such that $\beta(x_i)=m_i$, for $i\in I$. We also have that $f\circ\beta=\alpha\circ p$, because these two maps coincide on $B$.

Now consider the canonical injection $j\colon P\to P\oplus Q$; then $$ f_*(\beta\circ j)=f\circ\beta\circ j=\alpha\circ p\circ j=\alpha $$ because $p\circ j$ is the identity.