TheoremStatement: Literature-sourcedProof: Literature-sourcedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Covariant and contravariant $Hom$ are left exact

Statement

Let $X$ be a left $R$ -module.

If $0 \to A \overset{u}{\to} B \overset{v}{\to} C$ is exact, then $0 \to {Hom}_{R} (X, A) \overset{u_{*}}{\to} {Hom}_{R} (X, B) \overset{v_{*}}{\to} {Hom}_{R} (X, C)$ is exact.
If $A \overset{u}{\to} B \overset{p}{\to} C \to 0$ is exact, then $0 \to {Hom}_{R} (C, X) \overset{p^{*}}{\to} {Hom}_{R} (B, X) \overset{u^{*}}{\to} {Hom}_{R} (A, X)$ is exact.

Thus covariant and contravariant $Hom$ are left exact.

Facts & Assumptions

Given: The two exact sequences in the statement and a left $R$ -module $X$ .

[F1]

Postcomposition and precomposition define the displayed homomorphisms on Hom groups (The abelian group ${Hom}_{R} (M, N)$ and maps induced by pre- and postcomposition).

[F2]

Exactness means equality of the incoming image and outgoing kernel; the zero endpoints make $u$ injective in the first sequence and $p$ surjective in the second (Exact sequences and short exact sequences of modules).

[L1]

If a homomorphism $h : M \to P$ vanishes on a submodule $N$ , it factors uniquely through $M / N$ (A module homomorphism vanishing on $N$ factors uniquely through $M / N$ ).

Proof

technique · direct

1.1

If $u_{*} f = 0$ , then $u (f (x)) = 0$ for every $x$ , and injectivity of $u$ gives $f = 0$ ; hence $u_{*}$ is injective.

F1F2

1.2

Composability gives $v_{*} u_{*} = 0$ , so $im u_{*} \leq \ker v_{*}$ .

F1F2

1.3

If $g : X \to B$ satisfies $v_{*} g = 0$ , then $g (x) \in \ker v = im u$ for every $x$ . Injectivity of $u$ gives a unique $f (x) \in A$ with $u (f (x)) = g (x)$ ; uniqueness makes $f$ linear, so $g = u_{*} f$ .

F1F2

1.4

If $p^{*} h = 0$ , surjectivity of $p$ gives $h = 0$ , so $p^{*}$ is injective; and $u^{*} p^{*} = 0$ because $p \circ u = 0$ .

F1F2

1.5

If $g : B \to X$ satisfies $u^{*} g = g \circ u = 0$ , then $g$ vanishes on $im u = \ker p$ . By [L1] it factors uniquely through $B / \ker p$ , and since $p$ is surjective the rule $\overset{ˉ}{g} (p (b)) = g (b)$ gives the corresponding homomorphism $\overset{ˉ}{g} : C \to X$ with $g = \overset{ˉ}{g} \circ p = p^{*} (\overset{ˉ}{g})$ .

F1F2L1

2.1

Steps 1.1 to 1.3 prove the covariant sequence exact, and steps 1.4 and 1.5 prove the contravariant sequence exact.

step 1.1step 1.2step 1.3step 1.4step 1.5∎

Depends on

Used by

Hom_ℤ(-,ℤ) need not preserve surjections on the right Counterexample
Equivalent characterizations of injective modules Theorem
Equivalent characterizations of projective modules Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 28 results over 12 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

A. Kleshchev, Lectures on Abstract Algebra for Graduate Students, sections 3.6, 3.14, and 3.15 (standard reference, not scraped)
The Stacks Project, Algebra (standard reference, not scraped)
P. Hekmati, Homological Algebra, section 3.1 (standard reference, not scraped)

Covariant and contravariant Hom⁡Hom are left exact

Statement

Facts & Assumptions

Proof

Depends on

Used by

Dependency tree · next 3 levels

Sources

Covariant and contravariant $Hom$ are left exact