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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.

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Let

  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.

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  • $\begingroup$ 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
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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.

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