TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01

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Inverse limits preserve kernels

Statement

Let $(f_{n}) ⁣ : (M_{n}, φ_{n}) \to (N_{n}, ψ_{n})$ be a morphism of inverse systems of $R$ -modules. Then the kernel of the induced map $\underset{\leftarrow}{l i m} M_{n} ⟶ \underset{\leftarrow}{l i m} N_{n}$ is canonically isomorphic to $\underset{\leftarrow}{l i m} \ker (f_{n}) .$

Consequently, if $0 \to A_{n} \to B_{n} \to C_{n}$ is an exact sequence of inverse systems, then $0 \to \underset{\leftarrow}{l i m} A_{n} \to \underset{\leftarrow}{l i m} B_{n} \to \underset{\leftarrow}{l i m} C_{n}$ is exact.

Facts & Assumptions

Given: A morphism of inverse systems $(f_{n}) ⁣ : (M_{n}, φ_{n}) \to (N_{n}, ψ_{n})$ .

[L1]

The inverse limit consists of the compatible elements in the product system (Inverse systems and inverse limits of modules).

[L2]

The kernel of a homomorphism is the submodule of elements mapping to $0$ (Module homomorphism and isomorphism, kernel, image and cokernel).

[L3]

A map into an inverse limit is determined by its compatible coordinate maps (Universal property of an inverse limit of modules).

Proof

technique · direct

1.1L2given

For each $n$ , let $K_{n} : = \ker (f_{n})$ . Compatibility of the squares $ψ_{n} \circ f_{n} = f_{n - 1} \circ φ_{n}$ shows that $φ_{n} (K_{n}) \subseteq K_{n - 1}$ , so the kernels form an inverse system.

1.2L1L2

An element $(x_{n}) \in \underset{\leftarrow}{l i m} M_{n}$ lies in the kernel of $\underset{\leftarrow}{l i m} M_{n} \to \underset{\leftarrow}{l i m} N_{n}$ exactly when $f_{n} (x_{n}) = 0$ for every $n$ . By [L2], that is equivalent to $x_{n} \in K_{n}$ for every $n$ . Together with the compatibility condition from [L1], this says precisely that $(x_{n}) \in \underset{\leftarrow}{l i m} K_{n}$ .

2.1step 1.2

Thus the underlying subsets of $\ker ⁣ (\underset{\leftarrow}{l i m} M_{n} \to \underset{\leftarrow}{l i m} N_{n}) and \underset{\leftarrow}{l i m} \ker (f_{n})$ coincide inside $\prod M_{n}$ , and the module structures also agree componentwise. Hence these two modules are canonically equal, in particular canonically isomorphic.

3.1step 2.1L3∎

For an exact sequence $0 \to A_{n} \to B_{n} \to C_{n}$ , exactness means $A_{n} = \ker (B_{n} \to C_{n})$ for every $n$ . Applying step 2.1 to the maps $B_{n} \to C_{n}$ gives $\ker ⁣ (\underset{\leftarrow}{l i m} B_{n} \to \underset{\leftarrow}{l i m} C_{n}) = \underset{\leftarrow}{l i m} A_{n},$ which is exactly left exactness.

Inverse limits preserve kernels

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources