Given $R,S$ rings with unity, let $F: \textbf{mod-R} \longrightarrow \textbf{mod-S}$ a covariant additive functor between the categories of right modules.

We say that $F$ is exact when it preserves short exact sequences, ie: given $0\rightarrow A \stackrel{f}{\to} B\stackrel{g}{\to} C\rightarrow 0$ exact, the sequence $0 \stackrel{}{\to} FA \stackrel{F(f)}{\to} FB \stackrel{F(g)}{\to} FC \stackrel{}{\to}0$ is exact.

I want to do the following exercise:

Prove that F is exact if, and only if, for all exact sequences $A\stackrel{f}{\to}B\stackrel{g}{\to}C$, the sequence $FA \stackrel{F(f)}{\to} FB \stackrel{F(g)}{\to} FC$ is exact.

What I tried:

1) I observed that a sequence $A\stackrel{f}{\to}B\stackrel{g}{\to}C$ is exact if, and only if, the sequence $0\stackrel{}{\to} ran(f)\stackrel{i}{\to}B\stackrel{\pi}{\to} B/ker(g)\stackrel{}{\to}0$ is exact.

2) Also, I noticed that if $F$ is exact then it preserves injections and surjections between modules.



  1. for all exact sequences $A\xrightarrow fB\xrightarrow gC$, the sequence $FA\xrightarrow{F(f)}FB\xrightarrow{F(g)}FC$ is exact;
  2. for all exact sequences $0\to A\xrightarrow fB\xrightarrow gC\to 0$, the sequence $0\to FA\xrightarrow{F(f)}FB\xrightarrow{F(g)}FC\to 0$ is exact.

Clearly 1. implies 2., because $0\to A\xrightarrow fB\xrightarrow gC\to 0$ is exact if and only if \begin{gather} 0\to A\xrightarrow fB\\ A\xrightarrow fB\xrightarrow gC\\ B\xrightarrow gC\to 0 \end{gather} are all exact.

Conversely, let $A\xrightarrow fB\xrightarrow gC$ be exact. Then $0\to\operatorname{Ker}(f)\to A\xrightarrow f\operatorname{Im}(f)\to 0$ is exact hence $0\to F\operatorname{Ker}(f)\to FA\xrightarrow{F(f)} F\operatorname{Im}(f)\to 0$ is exact, hence $\operatorname{Ker}(F(f))=F\operatorname{Ker}(f)$ and $\operatorname{Im}(F(f))=F\operatorname{Im}(f)$ and, similarly, $\operatorname{Ker}(F(g))=F\operatorname{Ker}(g)$ and $\operatorname{Im}(F(g))=F\operatorname{Im}(g)$. Consequently, \begin{align} \operatorname{Im}(F(f))&=F(\operatorname{Im}(f))\\ &=F(\operatorname{Ker}(g))\\ &=\operatorname{Ker}(F(g)) \end{align} hence $FA\xrightarrow{F(f)}FB\xrightarrow{F(g)}FC$ is exact as wanted.

  • $\begingroup$ I could not prove that $FIm(f) = Im(F(f))$ and the same for $Ker$. I proved that they are isomorphic. Now I am writing my own answer. Thank you to show me your idea to solve the problem. $\endgroup$ – Victor Ronchim May 7 '18 at 21:35

To make the notation clearer, we will denote $f^F := F(f)$ for a morphism $f$.

Suppose that $F$ is exact. Given any function $f:A\longrightarrow B$, let $i:Kerf \to A$ and $j: Imf\to B$ the inclusion maps and $\tilde{f}:A\to Imf$ such that $j\circ \tilde{f} = f$. Since the sequence $0 \to Kerf \stackrel{i}{\to} A \stackrel{\tilde{f}}{\to}Imf \to 0$ is exact, we have:

  1. $0 \to FKerf \stackrel{i^F}{\to} FA \stackrel{\tilde{f}^F}{\to}FImf \to 0$ is exact;
  2. $\tilde{f}^F$ is surjective, i.e. $FImf= \tilde{f}^F[FA]$;
  3. $Imf^F = Im(j\circ\tilde{f})^F = Im(j^F\circ \tilde{f}^F) = j^F[\tilde{f}^F[FA]]= j^F[F Imf]$;
  4. Since $j^F$ is injective and $j^F\circ\tilde{f}^F= f^F$, then $Kerf^F = Ker\tilde{f}^F$;
  5. $Kerf^F = Ker\tilde{f}^F= Im\; i^F = i^F[FKerf]$.

From the fact that $F$ preserves injections, we have that $Kerf^F\cong FKerf$ and $Imf^F \cong FImf$.

Now, given $A\stackrel{f}{\to}B\stackrel{g}{\to}C$ exact, let $i=j$ the inclusion of $Kerg=Imf$ in $B$. We have: $$ Im f^F = i^F[F Imf] = i^F[FKerg] = Ker g^F.$$

Then, the sequence $FA \stackrel{f^F}{\to}FB \stackrel{g^F}{\to}FC$ is exact.

Conversely, note that a sequence $0 \to A\stackrel{f}{\to} B\stackrel{g}{\to} C\to 0$ is exact if it is exact in each term. Hence, $0 \to FA\stackrel{f^F}{\to} FB\stackrel{g^F}{\to} FC\to 0$ is exact because it is exact in each three lenght sequence.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.