Exact sequences of maps I have a question about exact sequences. If I have a sequence of the form $$0 \rightarrow A \xrightarrow{f}B \xrightarrow{g}C \rightarrow 0 $$ I know that for the sequence to be exact, we need that $Im(f) = Ker(g)$. However, do we need that $f$ be injective and $g$ be surjective? Or not?
Also, for the sequence $$ A \xrightarrow{f'} B \xrightarrow{g'}C $$ to be exact, besides having $Im(f') = Ker(g')$, do we need $f'$ to be injective or $g'$ to be surjective?
 A: For the first sequence, the answer is ‘yes’, because a $5$-term exact sequence is exact if and only if the $3$-term exact sequences:


*

*$0\to A\to B$,

*$A\to B\to C$ and

*$B\to C\to 0$


are exact. This translates exactly as you mention.
Obviously, the answer is ‘no’ for the second exact sequence.
A: A sequence of modules (or groups, rings, etc.) $A \xrightarrow{f} B \xrightarrow{g}C$ is exact if $\text{img}(f) = \ker(g)$.  A longer sequence
$$
\cdots \to A_{i-1} \xrightarrow{f_{i-1}} A_i \xrightarrow{f_i} A_{i+1} \xrightarrow{f_{i+1}} A_{i+2} \to \cdots
$$
is exact if it is exact at each $A_i$, i.e., if $\text{img}(f_i) = \ker(f_{i+1})$ for all $i$.
Now let's consider the $5$-term sequence
$$
0 \rightarrow A \xrightarrow{\varphi} B \xrightarrow{\psi} C \rightarrow 0 \, .
$$
Note that the image of the (unnamed) map $0 \to A$ is $0$, as the only map with domain $0$ is the $0$ morphism.  Similarly, the kernel of the map $C \to 0$ must be all of $C$.  So say that this sequence is exact means that it is exact at $A, B$ and $C$, i.e.,
(1) $\ker(\varphi) = \text{img}(0 \to A) = 0$
(2) $\ker(\psi) = \text{img}(\varphi)$
(3) $\text{img}(\psi) = \ker(C \to 0) = C$
Note that (1) means that $\varphi$ is injective and (3) means $\psi$ is surjective.
