Left-Exact Functors

We saw (in theorem 1 here) that the localization functor $M \mapsto S^{-1}M$ is exact, which gave us a whole slew of nice properties, including preservation of submodules, quotient modules, finite intersection/sum, etc. However, exactness is often too much to ask for.

Throughout this article, A and B are fixed rings and $F: A\text{-}\mathbf{Mod} \to B\text{-}\mathbf{Mod}$ is a covariant additive functor.

Definition.

We say F is left-exact if it takes a short exact sequence of A-modules

$0 \longrightarrow N \stackrel f\longrightarrow M \stackrel g\longrightarrow P \longrightarrow 0$

to an exact sequence of B-modules

$0 \longrightarrow F(N) \stackrel {F(f)}\longrightarrow F(M) \stackrel {F(g)} \longrightarrow F(P).$

Immediately we can weaken the condition.

Lemma 1.

F is left-exact if and only if it takes an exact sequence of A-modules

$0 \longrightarrow N \stackrel f\longrightarrow M \stackrel g\longrightarrow P$

to an exact sequence of B-modules

$0 \longrightarrow F(N) \stackrel {F(f)}\longrightarrow F(M) \stackrel {F(g)} \longrightarrow F(P).$

Proof

(⇐) is obvious. For (⇒) given an exact $0 \to N \stackrel f\to M \stackrel g\to P$ , we extend it to an exact $0 \to N \to M \to P \to \mathrm{coker } g \to 0$ . [Recall that $\mathrm{coker } g = P/\mathrm{im } g$ is the cokernel of g.] We can split this exact sequence into short exact ones:

$\begin{aligned} 0 \to N \stackrel f\to M \to \mathrm{im } g \to 0 &\implies 0 \to F(N) \stackrel{F(f)}\to F(M) \to F(\mathrm{im } g)\\ 0 \to \mathrm{im } g \to P \to \mathrm{coker } g \to 0 &\implies 0 \to F(\mathrm{im } g) \to F(P) \to F(\mathrm{coker } g).\end{aligned}$

which gives an exact sequence $0\to F(N) \to F(M) \to F(P)$ as desired. ♦

Note

Left-exact functors are not as nice as exact ones but we still get some useful results out of them. For example, if $f : N\to M$ is injective, so is $F(f) : F(N) \to F(M)$ so for a submodule $N\subseteq M$ , we can consider $F(N)$ as a submodule of $F(M)$ .

Also, for any $f:N\to M$ , since $0\to \mathrm{ker } f \to N \to M$ is exact, applying F gives an exact $0 \to F(\mathrm{ker} f) \to F(N) \stackrel{F(f)}\to F(M)$ and thus $F(\mathrm{ker} f) = \mathrm{ker } F(f)$ .

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Hom Functors Are Left-Exact

The main result we wish to show is

Proposition 1.

For any A-module M, the functor $\mathrm{Hom}_A(M, -)$ is a left-exact functor.

Proof

Take the exact sequence $0 \to N_1 \stackrel f\to N_2 \stackrel g\to N_3$ . We need to show that

$0 \longrightarrow \mathrm{Hom}_A(M, N_1) \stackrel {f_*}\longrightarrow \mathrm{Hom}_A(M, N_2) \stackrel{g_*}\longrightarrow \mathrm{Hom}_A(M, N_3)$

is exact. It is easy to show $f_*$ is injective (easy exercise). Next, we have

$g_*\circ f_* = (g\circ f)_* = 0 \implies \mathrm{im } f_* \subseteq \mathrm{ker } g_*.$

Conversely suppose $h:M\to N_2$ satisfies $g_*(h) = g\circ h = 0$ . Then $\mathrm{im } h \subseteq \mathrm{ker } g = \mathrm{im } f$ . Since f is injective it follows that for each $m\in M$ , we have $h(m) = f(x)$ for a unique $x\in N_1$ . This map $h' : M \to N_1, m\mapsto x$ is clearly A-linear so $h = f\circ h' = f_*(h')$ . ♦

Here is one application of this result.

Corollary 1.

Let $a\in A$ ; for each A-module, let $M[a] = \{m \in M : am = 0\}$ . Then

$0 \longrightarrow N \stackrel f \longrightarrow M \stackrel g \longrightarrow P \text{ exact } \implies 0 \longrightarrow N[a] \stackrel f\longrightarrow M[a] \stackrel g \longrightarrow P[a] \text{ exact.}$

Proof

Indeed the functor $M\mapsto M[a]$ is naturally isomorphic to $\mathrm{Hom}_A(A/(a), -)$ . Now apply the above. ♦

Exercise A

Find a surjective $M\to P$ for which $M[a] \to P[a]$ is not surjective. This shows that the functor $\mathrm{Hom}_A(M, -)$ is not exact in general.

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Converse Statement

Next we have the following converse.

Proposition 2.

Suppose $f:N_1 \to N_2$ and $g:N_2\to N_3$ are A-linear maps such that for any A-module M,

$0\longrightarrow \mathrm{Hom}_A(M, N_1) \stackrel {f_*} \longrightarrow \mathrm{Hom}_A(M, N_2) \stackrel {g_*} \longrightarrow \mathrm{Hom}_A(M, N_3)$

is exact. Then $0 \to N_1 \stackrel f \to N_2 \stackrel g \to N_3$ is exact.

Proof

First we show f is injective. Let $M =\mathrm{ker } f$ so that $0 \to M \to N_1 \stackrel {f} \to N_2$ is exact. By left-exactness of Hom we have an exact:

$0 \to {\mathrm{Hom}_A(M, M)} \to \mathrm{Hom}_A(M, N_1) \stackrel {f_*} \to \mathrm{Hom}_A(M, N_2)$ .

Since $f_*$ is injective, $\mathrm{Hom}_A(M, M) = 0$ and in particular $1_M = 0\implies M=0$ .

Next we show $g\circ f = 0$ . Setting $M = N_1$ gives:

$0 = g_* f_* (\mathrm{Hom}_A(N_1, N_1)) = (g\circ f)_*(\mathrm{Hom}_A(N_1, N_1))$

so $g\circ f = (g\circ f)_* (1_{N_1}) = 0.$ This gives $\mathrm{im } f \subseteq \mathrm{ker } g$ .

Finally we set $M = \mathrm{ker } g$ . The following sequence:

$0\longrightarrow \mathrm{Hom}_A(M, N_1) \stackrel {f_*} \longrightarrow \mathrm{Hom}_A(M, N_2) \stackrel {g_*} \longrightarrow \mathrm{Hom}_A(M, N_3)$

is exact. The inclusion $i : M\hookrightarrow N_2$ gives $g_*(i) = g\circ i = 0$ and so $i = f_*(j) = f\circ j$ for some $j : M \to N_1$ . Then $M = \mathrm{im } i \subseteq \mathrm{im } f$ . ♦

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Another Left-Exactness

Now suppose $F: A\text{-}\mathbf{Mod} \to B\text{-}\mathbf{Mod}$ is a contravariant additive functor.

Definition.

We say F is left-exact if it takes a short exact sequence of A-modules

$0 \longrightarrow N \stackrel f\longrightarrow M \stackrel g\longrightarrow P \longrightarrow 0$

to an exact sequence of B-modules

$0 \longrightarrow F(P) \stackrel {F(g)}\longrightarrow F(M) \stackrel {F(f)} \longrightarrow F(N).$

Again we have:

Lemma 2.

F is left-exact if and only if it takes an exact sequence of A-modules

$N \stackrel f\longrightarrow M \stackrel g\longrightarrow P \longrightarrow 0$

to an exact sequence of B-modules

$0 \longrightarrow F(P) \stackrel {F(g)}\longrightarrow F(M) \stackrel {F(f)} \longrightarrow F(N).$

Proof

Exercise. ♦

Now we show, as before:

Proposition 3.

For any A-module M, the functor $\mathrm{Hom}_A(-, M)$ is a left-exact functor.

Proof

Take the exact sequence $N_1 \stackrel f\to N_2 \stackrel g\to N_3 \to 0$ . We need to show that

$0 \longrightarrow \mathrm{Hom}_A(N_3, M) \stackrel {g^*}\longrightarrow \mathrm{Hom}_A(N_2, M) \stackrel{f^*}\longrightarrow \mathrm{Hom}_A(M, N_1)$

is exact for any A-module M. Injectivity of $g^*$ is obvious. Also $f^* \circ g^* = (g\circ f)^* = 0$ so $\mathrm{im } g^* \subseteq \mathrm{ker } f^*$ .

It remains to show $\mathrm{ker } f^* \subseteq \mathrm{im } g^*$ . Pick $h: N_2 \to M$ such that $f^*(h) = h\circ f = 0$ so that $\mathrm{ker } g = \mathrm{im } f \subseteq \mathrm{ker } h$ . Hence h factors through $N_2 / \mathrm{ker } g \to M$ . Since $N_2 / \mathrm{ker } g \cong N_3$ , we have $h = h'\circ g = g^*(h'')$ for some $h' : N_3 \to M$ . ♦

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Another Converse Statement

Finally, the reader should expect the following.

Proposition 4.

Suppose $f:N_1 \to N_2$ and $g:N_2\to N_3$ are A-linear maps such that for any A-module M,

$0\longrightarrow \mathrm{Hom}_A(N_3, M) \stackrel {g^*} \longrightarrow \mathrm{Hom}_A(N_2, M) \stackrel {f^*} \longrightarrow \mathrm{Hom}_A(N_1, M)$

is exact. Then $N_1 \stackrel f \to N_2 \stackrel g \to N_3 \to 0$ is exact.

Proof

We leave it as an exercise to show: g is surjective, $g\circ f = 0$ .

It remains to show $\mathrm{ker } g \subseteq \mathrm{im } f$ ; let $M = \mathrm{coker } f$ and take the exact sequence

$0\longrightarrow \mathrm{Hom}_A(N_3, M) \stackrel {g^*} \longrightarrow \mathrm{Hom}_A(N_2, M) \stackrel {f^*} \longrightarrow \mathrm{Hom}_A(N_1, M)$

If $\pi : N_2 \to M$ is the canonical map, $f^*(\pi) = \pi\circ f = 0$ so we have $\pi = g^*(j) = j\circ g$ for some $j : N_3 \to M$ . Thus $\mathrm{ker } g \subseteq \mathrm{ker } \pi = \mathrm{im } f$ . ♦

As an application, let us prove the following.

Corollary 2.

Let $\mathfrak a \subseteq A$ be an ideal. If $N \to M \to P \to 0$ is an exact sequence of A-modules, then we get an exact sequence of $(A/\mathfrak a)$ -modules:

$N/\mathfrak a N \longrightarrow M/\mathfrak a M \longrightarrow P/\mathfrak a P \longrightarrow 0.$

Proof

Let $B = A/\mathfrak a$ . It suffices to show that for any B-module Q, the sequence

$0 \longrightarrow \mathrm{Hom}_B(P/\mathfrak a P, Q) \longrightarrow \mathrm{Hom}_B(M/\mathfrak a M, Q) \longrightarrow \mathrm{Hom}_B(N/\mathfrak a N, Q)$

is exact. But since $M\mapsto M/\mathfrak a M$ gives the B-module induced from M, the above sequence is naturally isomorphic to:

$0 \longrightarrow \mathrm{Hom}_A(P, Q) \longrightarrow \mathrm{Hom}_A (M, Q) \longrightarrow \mathrm{Hom}_A(N, Q)$

which we know is exact because the Hom functor is left-exact. ♦

This leads to the following definition.

Definition.

Suppose $F: A\text{-}\mathbf{Mod} \to B\text{-}\mathbf{Mod}$ is a covariant additive functor. We say F is right-exact if it takes a short exact sequence of A-modules

$0 \longrightarrow N \stackrel f\longrightarrow M \stackrel g\longrightarrow P \longrightarrow 0$

to an exact sequence of B-modules

$F(N) \stackrel {F(f)}\longrightarrow F(M) \stackrel {F(g)} \longrightarrow F(P) \longrightarrow 0.$

Exercise B

Prove that F is right-exact if and only if it takes an exact sequence of A-modules $N \to M \to P \to 0$ to an exact sequence of B-modules $F(N) \to F(M) \to F(P) \to 0.$

Thus, we have shown that the functor

$A\text{-}\mathbf{Mod} \longrightarrow (A/\mathfrak a)\text{-}\mathbf{Mod}, \quad M \mapsto M/\mathfrak aM$

is right-exact. This will be generalized in future chapters.

Exercise C

Prove that the functor $M\mapsto M/\mathfrak a M$ is not exact in general.
Use a direct proof to show $M\mapsto M/\mathfrak a M$ is right-exact.

In summary, we have covered the following concepts:

exact_functor_types

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Commutative Algebra 26

Left-Exact Functors

Hom Functors Are Left-Exact

Converse Statement

Another Left-Exactness

Another Converse Statement

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Commutative Algebra 26

Left-Exact Functors

Hom Functors Are Left-Exact

Converse Statement

Another Left-Exactness

Another Converse Statement

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