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.


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: $0 \rightarrow F(A) \xrightarrow{F(\sigma)} F(B) \xrightarrow{F(\pi)} F(C)$


Of course if $F$ is a left exact functor, then it'll preserve monic.


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).


  • 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,

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    $\begingroup$ Remark: The quantors in the definition are wrong. $F$ is given, not the short exact sequence. $F$ is called left exact iff for every (!) short exact sequence, it gets transformed by $F$ to a left exact sequence. $\endgroup$ – Martin Brandenburg May 15 '13 at 12:48
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    $\begingroup$ I found that book to be very enjoyable, btw. Actually I can't remember if I read that one or his Introduction to homological algebra... they were very similar. Both good though. $\endgroup$ – rschwieb May 15 '13 at 13:08
  • $\begingroup$ @MartinBrandenburg Yes, thanks very much for pointing that out, I do make some mistakes reformulating the author's wording. My bad. $\endgroup$ – user49685 May 15 '13 at 14:23

I think you got to the point beginning with the sentence "Consider the exact sequence..." You can probably omit the work before that, although there is no mistake there. After that sentence, you get that the sequence is exact at $F(A)$ for free.

The mathematical leap was that you took the given and recognized that there is a relevant short exact sequence related to the exact sequence you were given, and you could apply the definition of "left exact functor" to that. To wrap things up, you showed that nothing was lost in the leap, and that you could pass back to the original sequence's maps.


Your proof is OK. It will become straight forward with a little bit more experience ;). You have done the only possible thing: Somehow extend the left exact sequence to a short exact sequence, using the image in order to make the last map surjective. By the way, I have given the same proof in SE/292037.

  • $\begingroup$ Thank you very much for correcting my wording. I'll try to be more careful in the future. Thanks man. :* $\endgroup$ – user49685 May 15 '13 at 14:28

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