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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Universal property of a direct sum of modules

Statement

Let $(M_{i})_{i \in I}$ be left $R$ -modules and $N$ a left $R$ -module. For every family of homomorphisms $f_{i} : M_{i} \to N$ , there is a unique homomorphism $f : ⨁_{i \in I} M_{i} ⟶ N$ such that $f \circ ȷ_{i} = f_{i}$ for every $i$ . It is given by $f ((m_{i})) = \sum_{i \in supp (m)} f_{i} (m_{i}) .$ For $I = \emptyset$ , this is the unique map $0 \to N$ .

Facts & Assumptions

Given: A family $(M_{i})_{i \in I}$ of left $R$ -modules, a left $R$ -module $N$ , and homomorphisms $f_{i} : M_{i} \to N$ .

[F1]

Elements of $⨁_{i} M_{i}$ have finite support, and $ȷ_{i}$ is the coordinate inclusion (The direct sum of an indexed family of modules).

[F2]

A module homomorphism preserves addition and scalar multiplication (Module homomorphism and isomorphism, kernel, image and cokernel).

Proof

technique · constructive

1.1

Define $f ((m_{i})) : = \sum_{i \in supp (m)} f_{i} (m_{i})$ ; the sum is finite by [F1], and padding it by zero terms shows it is independent of the chosen finite set containing the support.

F1construct

2.1

Addition and scalar multiplication may be checked on the finite union of the relevant supports, so [F2] gives $f (m + m^{'}) = f (m) + f (m^{'})$ and $f (r m) = r f (m)$ .

step 1.1F1F2

2.2

For $x \in M_{i}$ , step 1.1 gives $f (ȷ_{i} (x)) = f_{i} (x)$ , so $f \circ ȷ_{i} = f_{i}$ .

step 1.1F1

2.3

Every $m = (m_{i})$ equals the finite sum $\sum_{i \in supp (m)} ȷ_{i} (m_{i})$ . Hence any homomorphism $g$ satisfying $g \circ ȷ_{i} = f_{i}$ has $g (m) = \sum_{i} f_{i} (m_{i}) = f (m)$ and therefore equals $f$ .

step 1.1F1F2

3.1

If $I = \emptyset$ , the direct sum is $0$ by [F1], the formula is the empty sum, and the construction and uniqueness still apply. Thus the universal property holds for every index set.

step 1.1step 2.1step 2.2step 2.3F1discharge-construct∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 12 results over 6 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)