In Atiyah Macdonald, "Introduction to commutative Algebra" it says:
Proposition 2.9:i) Let $M' \xrightarrow[]{u}M \xrightarrow[]{v} M'' \rightarrow 0$
be a sequence of A-modules and homomorphisms. Then the sequence (4) is exact $\iff$ for all A-modules N, the sequence
$0\rightarrow Hom(M'',N)\xrightarrow[]{\bar{v}}Hom(M,N)\xrightarrow[]{\bar{u}} Hom(M',N)$ is exact.
Now my question: What does $Hom(M'',N)$ mean? I have looked in the text and on wikipedia and neither of them give me a straightforward answer. Part of me wants to say that it is a (group,category,module…something) which contains the homomorphisms from $M''$ to $N$.However if this was the case what would the maps $\bar{u}$ and $\bar{v}$ be?
Please note my knowledge of categories is not crash hot so please try and keep categorical comments as simple as possible.
Best Answer
$\operatorname{Hom}(M,N)$ refers to the set of $A$-module homomorphisms from $M$ to $N$. These form an abelian group under pointwise addition (define $f+g$ by $(f+g)(x)=f(x)+g(x)$), and if $A$ is commutative they in fact form an $A$-module by pointwise scalar multiplication (define $a\cdot f$ by $(af)(x)=a\cdot f(x)$).
Given an $A$-module homomorphism $g:M\to P$, there is an induced map $\bar{g}:\operatorname{Hom}(P,N)\to\operatorname{Hom}(M,N)$ that sends a homomorphism $f:P\to N$ to the composition $f\circ g:M\to N$. This map $\bar{g}$ is a homomorphism of abelian groups, and additionally a homomorphism of $A$-modules if $A$ is commutative.