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As a final exercise to VIII.1 in Algebra: Chapter $0$, we are asked to prove

If $\mathcal{F}\colon\operatorname{R-Mod}\to\operatorname{S-Mod}$ is a right-adjoint functor, then $\mathcal{F}$ is left-exact.

I am having some trouble proving this. If my overall strategy is right, then one necessary step is to show \begin{equation} \mathcal{F}(0)=0, \end{equation} where $0$ is the zero module. However I do not know how to do this.

Can someone give a hint? Thanks!

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3 Answers 3

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As Martin Brandenburg mentioned, this holds in a much more general context:

Let $C,D$ be categories and $F\colon C\rightarrow D$, $G\colon D\rightarrow C$ functors, such that $F$ is left adjoint to $G$. Then $F$ preserves all colimits and $G$ preserves all limits. Especially G preserves kernels and therefore is left-exact, whenever you can talk about exactness. (Dually: F preservers cokernels, so it's right-exact.)

This isn't hard to prove: You can prove by hand (checking the universal property), that covariant homfunctors $Hom(A,\_)$ preserve finite limits and then you can do the following by using the adjointness and the preservation of limits of hom-functors:

$$Hom(A,G(lim(X_i)))\cong Hom(F(A),lim(X_i))\cong lim(Hom(F(A),X_i))\cong lim(Hom(A,G(X_i)))\cong Hom(A,lim(G(X_i)))$$

All these isomorphisms are natural in A, so we get $Hom(\_,G(lim(X_i))\cong Hom(\_,lim(G(X_i)))$ and with the fact, that the Yoneda embedding is an embedding $G(lim(X_i))\cong lim(G(X_i))$. Dually this works for rightadjoints and colimits.

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    $\begingroup$ Not only finite limits. All limits. $\endgroup$
    – Martin Brandenburg
    Mar 23, 2013 at 17:02
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    $\begingroup$ This is proposition 3.3.6 in "Categories and sheaves" by Kashiwara and Schapira: functors admitting left adjoints are left-exact, and vice versa. Also the content of 3.3.3 is that a functor is left-exact iff it commutes with finite projective limits (when the category admits them). Note that they define left-exact functors $F : \mathcal{C}\to\mathcal{C}'$ as those such that, for every $U \in \mathcal{C}'$, the category $\mathcal{C}^U$ (whose objects are pairs of objects $X \in \mathcal{C}$ and morphisms $v : U \to F(X)$) is cofiltrant. $\endgroup$
    – user314
    Mar 23, 2013 at 17:43
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    $\begingroup$ @archipelago Right adjoints preserve all limits that exist. Not just finite, not just small, all of them. $\endgroup$
    – Zhen Lin
    Mar 23, 2013 at 17:45
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    $\begingroup$ @Adeel Generally we take left exact to mean preserves finite limits instead of the strong version you quote. $\endgroup$
    – Zhen Lin
    Mar 23, 2013 at 17:46
  • $\begingroup$ @ZhenLin Which reference would you recommend, that has this correct definition? $\endgroup$
    – user314
    Mar 23, 2013 at 17:49
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As other have stated, there are lots of good properties for adjoint functors. I will just prove your $F(0)=0$ statement. Let $G$ be left adjoint par to $F$. Then for all $S$-modules $X$ and $R$-modules $Y$ we have bijection $Hom_R(GX,Y) \simeq Hom_S(X,FY)$. Put $Y=0$ and $X=F0$. So we get:

$$Hom_S(F0,F0) \simeq Hom_R(GF0,0) \simeq \{0\}.$$

Now, if $F0\neq 0$, then there would be at least two different S-module-homomorphisms $F0 \to F0$ (namely, identity and zero-homomorphism), which would be a contradiction.

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To prove that a right-adjoint additive functor is left-exact, use the adjunction isomorphism to build a commutative rectangle from some exact sequence and then use the fact that the Yoneda embedding reflects exactness.

I'm unsure about the more general setting. Although in that case you would need to prove that $\mathcal{F}$ preserves finite limits, which includes preserving terminal objects, i.e. $\mathcal{F}(0)=0$.

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    $\begingroup$ Exact functors are defined between arbitrary categories. Left exact = preservation of finite limits, right exact = preservation of finite colimits, exact = left exact and right exact. $\endgroup$
    – Martin Brandenburg
    Mar 23, 2013 at 16:12
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    $\begingroup$ @MartinBrandenburg: I see. Although in the notes linked by the OP exact functors are defined as functors preserving exact sequences, and from here it looks like this implies your definition just for Ab-enriched, i.e. abelian, functors. Is this correct? $\endgroup$
    – A.P.
    Mar 23, 2013 at 16:19
  • $\begingroup$ Yeah, it implies the "exact-sequence-definition", as I mentioned it in my answer. Kernels are limits and cokernels are colimits, whenever you can talk about kernels and cokernels. (e.g. in abelian categories) $\endgroup$
    – archipelago
    Mar 23, 2013 at 17:23
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    $\begingroup$ It seems to me you have to first assume that the functor is additive, but in the exercise we only know it is right-adjoint. Is additivity necessary? $\endgroup$
    – Hui Yu
    Mar 24, 2013 at 9:23
  • $\begingroup$ Not in general. Please see archipelago's nice answer. $\endgroup$
    – A.P.
    Mar 24, 2013 at 11:12

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