# If a functor between categories of modules preserves injectivity and surjectivity, must it be exact?

Let $A$ and $B$ be commutative rings. Let $F$ be a functor from the category of $A$ modules to the category of $B$ modules. Suppose that $F$ preserves injectivity and surjectivity: whenever $f : X\rightarrow Y$ is an injective map of $A$-modules, we have that $F(f) : F(X)\rightarrow F(Y)$ is an injective map of $B$ modules, and similarly for a surjective map. In other words, $F$ is both left and right exact. Is $F$ necessarily exact, in the sense that it takes short exact sequences to short exact sequences?

It seems like all of the examples of non-exact functors one usually sees ($Hom$, $\otimes$, etc.) fail to preserve either injectivity or surjectivity, so I am wondering if there are any examples which preserve both.

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## 2 Answers

No. For example, the symmetric square functor $S^2(-) : \text{Vect} \to \text{Vect}$ preserves both injective and surjective maps, but it is not exact.

Your assertion that $F$ is left exact if $F$ preserves injective maps is false without the additional assumption that $F$ is additive (edit: and now that I think about it may be false even with this assumption...?). It is true that if $F$ is an additive functor which is both left and right exact, then $F$ is exact. But there are several equivalent definitions of left exactness for an additive functor (between abelian categories) which are no longer equivalent if $F$ is not additive.

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Can you give a more detailed description of the "symmetric square" functor? – user15464 Jan 3 '13 at 2:06
@user: it sends a vector space $V$ to the quotient of the tensor square $V \otimes V$ by the subspace of vectors of the form $v \otimes w - w \otimes v$. Feel free to substitute the tensor square functor if you're more comfortable with that. If $e_1, e_2, ...$ is a basis of $V$, then $\{ e_i e_j : i \le j \}$ is a basis of $S^2(V)$. – Qiaochu Yuan Jan 3 '13 at 2:08
Perhaps it's worth mentioning that a right exact functor between abelian categories is exact if and only if it preserves monomorphisms – which is probably where the flawed analogy comes from. – Zhen Lin Jan 3 '13 at 4:41
For a nice example of an additive functor between abelian categories which preserves monos but is not left exact, one can see Martin Brandenburg's answer here. – Ben Dyer Jul 11 '15 at 23:08

To amplify, let me point out that it is not even true that an additive functor that preserves epimorphisms/surjections also preserves cokernels. For example, let $\mathcal{C}$ be the full subcategory of $\textbf{Ab}$ spanned by the torsion-free abelian groups. This category, perhaps unexpectedly, is an additive category with kernels and cokernels – but is not an abelian category. Indeed, in $\mathcal{C}$, the cokernel of $2 \cdot {-} : \mathbb{Z} \to \mathbb{Z}$ is $0$, so the inclusion $\mathcal{C} \hookrightarrow \textbf{Ab}$ is an additive functor that preserves surjections but not cokernels (or even epimorphisms in general!). In fact, $\mathcal{C} \hookrightarrow \textbf{Ab}$ is even a right adjoint, and so is left exact in particular. (When I say left/right exact, I always mean a functor that preserves finite limits/colimits.)

However, what is true is that a left exact functor $F : \mathcal{A} \to \mathcal{B}$ that preserves (normal) epimorphisms will be exact, provided $\mathcal{A}$ and $\mathcal{B}$ are both abelian. Indeed, since $\mathcal{A}$ and $\mathcal{B}$ are additive, to show that $F$ is right exact it is enough to show that it is additive and preserves cokernels. Now, it is known that a functor $\mathcal{A} \to \mathcal{B}$ that preserves finite products is automatically additive; but a left exact functor preserves finite products, so is an additive functor in particular. Consider a sequence of morphisms $$0 \longrightarrow A' \longrightarrow A \longrightarrow A'' \longrightarrow 0$$ in $\mathcal{A}$, and suppose $A' \to A$ is kernel of $A \to A''$ and $A \to A''$ is the cokernel of $A' \to A$. Since $F$ preserves kernels, we get an exact sequence $$0 \longrightarrow F A' \longrightarrow F A \longrightarrow F A''$$ in $\mathcal{B}$, and since $A \to A''$ is a (normal) epimorphism, we can extend the above to a short exact sequence in $\mathcal{B}$: $$0 \longrightarrow F A' \longrightarrow F A \longrightarrow F A'' \longrightarrow 0$$ Thus, $F$ preserves cokernels of normal monomorphisms. In general, if we have a morphism $X \to A$ in $\mathcal{A}$, we can factor it as $X \to A' \to A$ where $X \to A'$ is the cokernel of the kernel of $X \to A$, and it is not hard to show that the cokernel of $A' \to A$ is also the cokernel of $X \to A$. Since $F$ preserves all kernels and also cokernels of (normal) monomorphisms, $F$ preserves this factorisation, and therefore the cokernel of $F A' \to F A$ is also the cokernel of $F X \to F A$. However, because $\mathcal{A}$ is an abelian category, $A' \to A$ itself is a (normal) monomorphism, so $F$ preserves its cokernel. Thus $F$ actually preserves all cokernels and is therefore right exact.

Dually, of course, a right exact functor between abelian categories is exact if and only if it preserves (normal) monomorphisms. This explains the classical fact that $M$ is flat if and only if $- \otimes_R M$ preserves injective homomorphisms: $- \otimes_R M$ is a left adjoint and so right exact in particular.

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What do you mean by your repeated parenthesis "(normal)"? The exactness of the last displayed short exact sequence ultimately relies on the axiom/lemma that an epimorphism in an abelian category is the cokernel of its kernel, right? – Martin Jan 3 '13 at 8:05
A normal mono/epimorphism is one that is a kernel/cokernel. An abelian category is precisely an additive category with kernels and cokernels in which all monomorphisms and all epimorphisms are normal. – Zhen Lin Jan 3 '13 at 8:29
Got it. Thanks. – Martin Jan 3 '13 at 8:35