I'm reading A first course of Homological Algebra by Northcott, and there is something that the author said it was straightforward. But for some reason, I just don't see the straightforwardness of it.
Definition
And let $F:\mathscr{C}_\Lambda \to \mathscr{C}_\Delta$; where $\mathscr{C}_\Lambda$; $\mathscr{C}_\Delta$ be the covariant functor between the category of $\Lambda-$left module, and $\Delta-$left module.
$F$ is said to be left exact iff for every exact sequence of left module: $0 \rightarrow A \xrightarrow{\sigma} B \xrightarrow{\pi} C \rightarrow 0$, the following induced sequence is also exact: $0 \rightarrow F(A) \xrightarrow{F(\sigma)} F(B) \xrightarrow{F(\pi)} F(C)$
Property
Of course if $F$ is a left exact functor, then it'll preserve monic.
Theorem
Suppose $F:\mathscr{C}_\Lambda \to \mathscr{C}_\Delta$ is a left exact covariant functor, and that $0 \rightarrow A \xrightarrow{\sigma} B \xrightarrow{\pi} C$ is exact in $\mathscr{C}_\Lambda$. Prove that $0 \rightarrow F(A) \xrightarrow{F(\sigma)} F(B) \xrightarrow{F(\pi)} F(C)$ is also exact in $\mathscr{C}_\Delta$.
Here's my proof, I'll be very glad if you guys can check my proof, as well as give me some hints to prove it in another easier way (if another way exists).
Proof
Since $F$ preserves monic, the sequence $0 \rightarrow F(A) \xrightarrow{F(\sigma)} F(B) \xrightarrow{F(\pi)} F(C)$ is exact at $F(A)$.
$\pi \sigma = 0 \Rightarrow 0 = F(\pi \sigma) = F(\pi)F(\sigma)$
What left is to prove $\ker F(\pi) \subset \mbox{im } F(\sigma)$
Consider the exact sequence: $0 \rightarrow A \xrightarrow{\sigma} B \xrightarrow{\pi} \mbox{im }\pi' \rightarrow 0$, where $\pi'$ is the same as $\pi$ with the codomain contracted to be $\mbox{im } \pi$ since $F$ is left exact, it means that: $0 \rightarrow F(A) \xrightarrow{F(\sigma)} F(B) \xrightarrow{F(\pi')} F(\mbox{im }\pi)$ is also exact.
Consider the following chain of homomorphism: $F(B) \xrightarrow{F(\pi')} F(\mbox{im }\pi) \xrightarrow{F(i)} F(C)$
Since $i$ is monic, $F(i)$ is also monic (I think so, but is it right?)
So $\ker F(\pi) = \ker [F(i)F(\pi')] = \ker F(\pi') = \mbox{im }F(\sigma)$.
Is my proof valid? And is there any shorter proof? As the author said that it's straightforward, and I don't really think my proof can be classified to be straightforward.
Thanks guys very much,
Have a good day,