# Abstract Nonsense

## Left Exact, Right Exact, and Exact Functors

Point of Post: In this post we discuss the notion of left exact, right exact, and exact functors between abelian categories–giving several equivalent definitions.

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Motivation

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We finally get back to the homological algebra side of things. What we would now like to discuss is just a thinly veiled continuation of our last post on continuous and cocontinuous functors. Namely, we are going to ask when a functor preserves/partially preserves the exactness of a given short exact sequence in some abelian category. This is of absolutely fundamental importance in homological algebra because as we have previously stated the degree to which a sequence fails to be exact is the reason we study homological algebra and thus it behooves us to figure out when applying a functor to an exact sequence introduces no new obstruction. In fact, a huge part of the homological algebra to come will be measuring how badly a certain type of functor fails to be exact.

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Left Exact, Right Exact, and Exact Functors

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Let us get the definitions out of the way.

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Let $\mathscr{A}$ and $\mathscr{B}$ be abelian categories. We call a functor $F:\mathscr{A}\to\mathscr{B}$ left exact if for every exact sequence

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$0\to X\xrightarrow{\alpha} Y\xrightarrow{\beta} Z$

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in $\mathscr{A}$ the sequence

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$0\to F(X)\xrightarrow{F(\alpha)}F(Y)\xrightarrow{F(\beta)}F(Z)$

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is exact in $\mathscr{B}$. We say that $F$ is right exact if every exact sequence $X\xrightarrow{\alpha}Y\xrightarrow{\beta}Z\to0$ in $\mathscr{A}$ goes to an exact sequence

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$F(X)\xrightarrow{F(\alpha)}F(Y)\xrightarrow{F(\beta)}F(Z)\to0$

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in $\mathscr{B}$. We call a functor $F$ exact if it is both left and right exact.

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For example, we have shown that $\otimes_R A$ is a right exact functor $R\text{-}\mathbf{Mod}\to R\text{-}\mathbf{Mod}$ for commutative rings $R$ and we have also shown that $\text{Hom}_R(A,\bullet)$ is a left exact exact functor $R\text{-}\mathbf{Mod}\to R\text{-}\mathbf{Mod}$ with the same conditions. As a different example we have shown that the direct limit functor $\varinjlim:\mathbf{DS}_\mathcal{A}(R\text{-}\mathbf{Mod})\to R\text{-}\mathbf{Mod}$ is exact when $\mathcal{A}$ is directed–or said using our more recently language, if $\mathcal{A}$ is a directed set then the colimit functor $\text{colim }:(R\text{-}\mathbf{Mod})^\mathcal{A}\to R\text{-}\mathbf{Mod}$ is exact.

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Now, there are some nice equivalent definitions of left/right exactness if we assume our functors are additive. Namely:

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Theorem: Let $F:\mathscr{A}\to\mathscr{B}$ be an additive functor. Then, the following are equivalent:

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\begin{aligned}&\mathbf{(1)}\quad F\text{ is left exact}\\&\mathbf{(2)}\quad F\text{ commutes with kernels}\\ &\mathbf{(3)}\quad F\text{ is finitely continuous}\\ &\mathbf{(4)}\quad F\text{ preserves left short exact sequences}\end{aligned}

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Proof:

$\mathbf{(1)}\implies\mathbf{(2)}$: Let $X\xrightarrow{\alpha}Y$ be an arbitrary arrow. We then have the exact sequence $\ker \alpha\xrightarrow{k} X\xrightarrow{\alpha}Y$ which applying $F$ gives the exact sequence $0\to F(\ker \alpha)\xrightarrow{F(k)} F(X)\xrightarrow{F(\alpha)}F(Y)$. By exactness we have a natural isomorphism $\text{im }F(k)\xrightarrow{\approx}\ker F(\alpha)$, but since $F(k)$ is a monomorphism it’s trivial $\text{im }F(k)$ is naturally isomorphic to $F(\ker \alpha)$. The rest follows from the fact that the category $\bullet\overset{\displaystyle \longrightarrow}{\longrightarrow}\bullet$ (which is the category that equalizers, and thus kernels are limits of) is connected and applying theorem 2.2 of this paper.

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$\mathbf{(2)}\implies\mathbf{(3)}$: Since $F$ is additive we know that it commutes with products, and so if it also commutes with kernels, and since all equalizers are just kernels in abelian categories we know that $F$ commutes with all finite limits (this follows from previous comment) and thus is finitely cocontinuous.

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$\mathbf{(3)}\implies\mathbf{(4)}$: Let $0\to X\xrightarrow{\alpha}Y\xrightarrow{\beta}Z\to0$ be a short exact sequence in $\mathscr{A}$. We need to verify that $F(\alpha)$ is a monomorphism and that $\text{im }F(\alpha)\cong\ker F(\beta)$. Now, to see that $F(\alpha)$ is a monomorphism we merely note that by assumption $\ker F(\alpha)\cong F(\ker\alpha)\cong F(0)=0$. One can proceed similarly for the other point of exactness.

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$\mathbf{(4)}\implies\mathbf{(1)}$: This one is trivial.  $\blacksquare$

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Of course, there is a dualized version of this for right exact functors where we “co” everything.

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This tells us that, in particular, things like the tensor functor $\otimes_R N$ where $R$ is commutative is right exact, since it is left adjoint to the Hom functor and so cocontinuous and so right exact. This also allows us to finally complete a long-standing promise. Namely, a long time ago we claimed that the inverse limit functor was left exact. This now follows immediately since this is just a limit functor which we know is a right adjoint to the diagonal functor, and so continuous, and so finitely continuous and so left exact. Cool, huh?

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References:

[1] Mac, Lane Saunders. Categories for the Working Mathematician. New York: Springer-Verlag, 1994. Print.

[2] Adámek, Jirí, Horst Herrlich, and George E. Strecker. Abstract and Concrete Categories: the Joy of Cats. New York: John Wiley & Sons, 1990. Print.

[3] Berrick, A. J., and M. E. Keating. Categories and Modules with K-theory in View. Cambridge, UK: Cambridge UP, 2000. Print.

[4] Freyd, Peter J. Abelian Categories. New York: Harper & Row, 1964. Print.

[5] Mitchell, Barry. Theory of Categories. New York: Academic, 1965. Print.

[6] Herrlich, Horst, and George E. Strecker. Category Theory: An Introduction. Lemgo: Heldermann, 2007. Print.

April 24, 2012 -