The meaning of functor $M \mapsto \mbox{Hom}_A(P,M)$ being exact

Question

(I'm currently studying Lang's text Algebra and it comes up on the page 137. Lang does not explicitly define this expression.)

Is the following understanding correct?

The function $M \mapsto \mbox{Hom}_A(P,M)$ is called exact if the following holds: A sequence $$ 0 \to M' \to M \to M'' $$ is exact if and only if (should I take out one of if and only if?) $$ 0 \to \mbox{Hom}_A(P,M') \to \mbox{Hom}_A(P,M) \to \mbox{Hom}_A(P,M'') $$.

Here we are dealing with modules and as you may already know, if the above most condition holds then $P$ is projective.

Answer 1

The functor $\newcommand{\Hom}{\operatorname{Hom}}M\mapsto\Hom_A(P,M)$ is left exact for any module $P$; this means that from exactness of $$ 0\to M'\to M\to M'' $$ you can always deduce exactness of $$ 0\to\Hom_A(P,M')\to\Hom_A(P,M)\to\Hom_A(P,M'') $$ The converse implication need not hold when $P$ is a projective module; indeed, the zero module is projective, so if your claim would be true, then any sequence of the form $0\to M'\to M\to M''$ would be exact, because $$ 0\to\Hom_A(0,M')\to\Hom_A(0,M)\to\Hom_A(0,M'') $$ is obviously exact.

Answer 2

Yes, you should remove the "only if". Now, for the sequences, you need to add another zero to the right hand side, so it reads: $$0 \rightarrow {M}' \rightarrow M \rightarrow {M}'' \rightarrow 0$$ which is then sent to $$0 \rightarrow Hom_A(P,{M})' \rightarrow Hom_A(P,M) \rightarrow Hom_A(P,{M}'') \rightarrow 0$$ As you put it, the functor is called left-exact, and if there were a single zero on the right, it would be right exact.