# Limits Are Left-Exact

By example 6 and proposition 2 in the previous article, one is inclined to conclude that taking the colimit in $\mathcal C = A\text{-}\mathbf{Mod}$ is a right-exact functor, but there is a rather huge issue here: the functors are between $\mathcal C$ and $\mathcal D = \mathcal C^J$, the category of diagrams in $\mathcal C$ while we only defined exactness of functors between categories of modules. The proper way to do this is to introduce the framework of abelian categories and extend our concept of additive functors and exact functors there. However, doing this will take us too far afield so we will prove it directly (which is, admittedly, a bit of a cop out).

Proposition 1.

Let J be an index category, and $D', D, D'' : J\to A\text{-}\mathbf{Mod}$ be diagrams of type J. For concreteness, write these diagrams as

$((N_i), (\beta_e^N : N_i \to N_j)), \ ((M_i), (\beta_e^M : M_i \to M_j)),\ ((P_i), (\beta_e^P : P_i \to P_j))$

where $i\in J$ and $e:i\to j$. Let $D'\to D \to D''$ be morphisms, written as a collection of $N_i\stackrel {\phi_i} \to M_i \stackrel {\psi_i} \to P_i$ over $i\in J$. Then

\left( \begin{aligned}0\to N_i\stackrel {\phi_i} \to M_i \stackrel {\psi_i} \to P_i\\ \text{exact for each } i\in J\end{aligned}\right) \implies 0 \to \lim N_i \to \lim M_i \to \lim P_i \text{ exact.}

\left( \begin{aligned}N_i\stackrel {\phi_i} \to M_i \stackrel {\psi_i} \to P_i \to 0\\ \text{exact for each } i\in J\end{aligned}\right) \implies \mathrm{colim} N_i \to \mathrm{colim} M_i \to \mathrm{colim} P_i \to 0\text{ exact.}

Note

In summary, taking the limit is left-exact while taking the colimit is right-exact.

Proof

We prove the second claim, leaving the first as an exercise. By proposition 1 here, $\mathrm{colim} M_i$ is concretely described as follows. Take the quotient of $\oplus_{i\in J} M_i$ by all $m_i - \beta_e^M(m_i)$, where $e:i\to j$ is an arrow in J, $m_i \in M_i$ and $\beta_e^M(m_i)\in M_j$ are identified with their images in $\oplus_i M_i$.

With this description, clearly $\mathrm{colim} M_i \to \mathrm{colim} P_i$ is surjective. Also, composing $\mathrm{colim} N_i \to \mathrm{colim} M_i \to \mathrm{colim} P_i$ is the zero map so $\mathrm{im}(\mathrm{colim} \phi_i) \subseteq \mathrm{ker}(\mathrm{colim} \psi_i)$. Now write $\phi : \oplus_i N_i \to \oplus_i M_i$ for $\oplus \phi_i$ and $\psi : \oplus_i M_i \to \oplus_i P_i$ for $\oplus \psi_i$.

Conversely, let $x\in \oplus_i M_i$ represent an element in the kernel of $\mathrm{colim} \psi_i$. Thus $\psi(x) \in\oplus_i P_i$ is a finite sum of $p_i - \beta_e^P(p_i)$. Since $\psi_i$ is surjective, we can write such a term as

$\psi_i(m_i') - \beta_e^P(\psi_i(m_i')) = \psi_i(m_i') - \psi_j(\beta_e^M(m_i')) = \psi(m_i' - \beta_e^M(m_i'))$

for some $m_i'\in M_i$. Since $\psi(x)$ is a finite sum of $\psi(m_i' - \beta_e^M(m_i'))$, we can replace x by another representative such that $\psi(x) = 0$. Then $x = \phi(y)$ for some $y\in \oplus N_i$. ♦

Neither the limit nor the colimit functor is exact in general. For the colimit case, consider the following commutative diagram of A-modules

where all maps $A\to A$ are identities. The rows are short exact sequences and the squares all commute, but taking the colimit of the columns gives

$0 \longrightarrow A^2 \longrightarrow A \longrightarrow 0 \longrightarrow 0$

which is not exact.

Exercise A

Find an example for the case of limits.

# Direct Limits

We will describe a special case where taking the colimit is exact.

Given a poset $(S, \le)$, we recall the category $\mathcal C(S)$ whose objects are elements of S, and between any $x,y\in S$, $|\mathrm{hom}(x, y)| \le 1$ with equality if and only if $x \le y$. Composition is the obvious one.

Definition.

A poset $(S, \le)$ is called a directed set if for any $a,b\in S$, there is a $c\in S$ such that $a\le c$ and $b\le c$.

In other words, a poset is directed if every finite set has an upper bound.

Definition.

If J is an index category obtained from $\mathcal C(S)$ for some directed set S, then a diagram in $\mathcal C$ of type J is called a directed system. The colimit of $(A_i)_{i\in J}$ is called the direct limit and denoted by

$\varinjlim_{i\in J} A_i$.

In other words, direct limit = colimit over directed set. We will abuse notation a little and regard J as the directed set itself.

To avoid set-theoretic difficulties, the directed set J is always assumed to be non-empty.

Example

In exercise C.3 here, for a multiplicative $S\subseteq A$ and A-module M, we have an isomorphism of A-modules

$\mathrm{colim}_{f\in S} M_f \cong M_S$

where $f\le g$ if g is a multiple of f. Since S is multiplicative, any {fg} has an upper bound fg. Hence $M_S$ is the direct limit of $M_f$ over $f\in S$:

$\varinjlim_{f\in S} M_f \cong M_S$.

Similarly, we have the following direct limit in the category of rings:

$\varinjlim_{f\in S} A_f \cong A_S$.

Next we will discuss the general direct limit in the categories A-Mod and Ring.

# Direct Limit of Modules

Let A be a fixed ring; the following holds for direct limits in the category of A-modules.

Proposition 2.

Suppose $((M_i)_{i\in J}, (\beta_{ij})_{i\le j})$ is a directed system of A-modules over a directed set J. Let

$M = \varinjlim_{i\in J} M_i$, with canonical $\epsilon_i : M_i \to M$ for each $i\in J$.

Then for each $m\in M$, there exists an $i\in J$ and $m_i \in M_i$ such that $\epsilon_i(m_i) = m$.

Also if $m_i\in M_i$ satisfies $\epsilon_i(m_i) = 0$, then there exists $j\ge i$ such that $\beta_{ij}(m_i) = 0 \in M_j$.

Note

The philosophy is that “whatever happens in the direct limit happens in $M_j$ for some sufficiently large index j“.

Proof

By proposition 1 here, the colimit M is described concretely by taking the quotient of $P = \oplus_{i\in J} M_i$ (with canonical $\nu_i : M_i \to P$) by relations of the form

$\nu_k(m_k) - \nu_l\beta_{kl}(m_k),\ m_k \in M_i,\ k\le l\ (k, l\in J).$

Hence any $m\in M$ can be written as $\epsilon_{i_1}(m_1) + \ldots + \epsilon_{i_N}(m_N)$ for $m_1 \in M_{i_1}, \ldots, m_N \in M_{i_N}$. But J is a directed set, so we can pick index $j\in J$ such that $j\ge i_1, \ldots, i_N$; then

$m = \epsilon_j(m_j)$, where $m_j = \beta_{i_1 k}(m_1) + \ldots + \beta_{i_N k}(m_N)$,

proving the first claim.

For the second claim, if $\epsilon_i(m_i) = 0$ then $\nu_i(m_i) \in \oplus_i M_i$ is a finite sum of the above relations. Pick an index $j\in J$ larger than i and all indices kl in the sum; then $\beta_{ij}(m_i)$ is the sum of the images of these relations in $M_k$. But each such relation has image $\beta_{kj}(m_k) - \beta_{lj}\beta_{kl}(m_k) = 0$ in $M_k$, so $\beta_{ij}(m_i) = 0$ as desired. ♦

Corollary 1.

If $((M_i), (\beta_{ij}))$ is a directed system of A-modules such that $\beta_{ij}$ are all injective, then

$\epsilon_j : M_j \longrightarrow \varinjlim_i M_i$

is also injective for each $j\in J$.

Finally we have:

Proposition 3.

Let $((M_i), (\beta_{ij}))$ and $((N_i), (\gamma_{ij}))$ be directed systems of A-modules and $\phi_i : M_i \to N_i$ be a morphism of the directed systems, i.e. for any $i\le j$, we have $\gamma_{ij} \circ \phi_i = \phi_j \circ \beta_{ij} : M_i \to N_j$.

If each $\phi_i$ is injective, so is $\phi : \varinjlim M_i \to \varinjlim N_i$.

Since taking the colimit is right-exact by proposition 1, we see that taking the direct limit is exact.

Proof

Write $\epsilon_i^M : M_i \to \varinjlim M_i$ and $\epsilon_i^N : N_i \to \varinjlim N_i$ for the canonical maps.

Suppose $\phi(m) = 0$ for $m \in \varinjlim M_i$. By proposition 2, we have $m = \epsilon_i^M(m_i)$ for some $m_i \in M_i$; then

$0 = \phi(m) = \phi(\epsilon_i^M(m_i)) = \epsilon_i^N(\phi_i(m_i))$

so by proposition 2 again, there exists $j\ge i$ such that $\gamma_{ij}(\phi_i(m_i)) = 0$, so $\phi_j(\beta_{ij}(m_i)) = 0$. Since $\phi_j$ is injective we have $\beta_{ij}(m_i) = 0$ so $m = \epsilon_i^M(m_i) = \epsilon_j^M\beta_{ij}(m_i) = 0$. ♦

Exercise B

Describe the direct limit of sets $(S_i)$ over J. State and prove an analogue of proposition 2.

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### 2 Responses to Commutative Algebra 51

1. Vanya says:

In proposition are the maps $\latex \phi_i : M_i \to N_i$ instead of $\phi_i : N_i \to M_i$?

• limsup says:

Thanks corrected the bug(s) in prop 3.