My understood definition: $0 \to A\stackrel{\psi} \to B\stackrel{\phi} \to C \to 0$ is a short exact sequence of modules iff $\psi$ is injective, $\phi$ is surjective, and $\ker \phi = \psi(A)$, which is equivalent to $B/A \approx C$. Is that correct?
And also how do you arrive at $A\stackrel{\psi} \to B\stackrel{\phi}\to C$ is exact iff $0 \to \psi(A) \to B \to B/\ker \phi \to 0$ is short exact?
Best Answer
For the first part, the general definition is that a sequence $\cdots \to L \xrightarrow{f} M \xrightarrow{g} N \to \cdots $ is exact at $M$ if and only if the image of $f$ is the kernel of $g$.
A short exact sequence is an sequence consisting of five terms whose endpoints are zero, and is exact at every (internal) point. i.e.
$$ 0 \to A \xrightarrow{f} B \xrightarrow{g} C \to 0 $$
is exact at $A$, at $B$ and at $C$. Exactness at $A$ means that the kernel of $f$ is the image of the zero map: i.e. it is $\{ 0 \}$, so exactness at $A$ is equivalent to $f$ being injective. Similarly for exactness at $C$.
Usually it doesn't even make sense to say $B/A$, because $A$ usually isn't a submodule of $B$! However, there is equivalence in the sense that there is an isomorphism of short exact sequences:
$$ \begin{matrix} 0 &\to & A &\xrightarrow{f}& B &\xrightarrow{g}& C &\to& 0 \\ & & \ \downarrow^f & & \ \downarrow^1 & & \downarrow \\ 0 &\to & f(A) &\xrightarrow{i}& B &\xrightarrow{\pi}& B / f(A) &\to& 0 \end{matrix} $$
where both rows are short exact sequences, $i$ and $\pi$ are the inclusion and projection morphisms, all three vertical morphisms are isomorphisms, and the diagram is commutative. (meaning, e.g., if you take all paths from the top $A$ to the bottom $B$ and compose the arrows, you get the same result: $i \circ f = 1 \circ f$)
One can view that the situation as per the definition of quotient module is somewhat deficient, and the idea of a short exact sequence is what one really should be thinking about.