TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Algebra homomorphisms are closed under identities and composition

Statement

For a monad $T$ , identity morphisms are $T$ -algebra homomorphisms, and the composite of two $T$ -algebra homomorphisms is a $T$ -algebra homomorphism. These operations inherit associativity and identity laws from the base category.

Facts & Assumptions

Given: $T$ -algebras and their homomorphisms as in Algebra and algebra homomorphism for a monad.

Proof

technique · direct

1.1given

For an algebra $(A, a)$ , functoriality gives $T (1_{A}) = 1_{T A}$ , so $1_{A} \circ a = a = a \circ T (1_{A})$ and $1_{A}$ is an algebra homomorphism.

2.1givenstep 1.1

If $f : (A, a) \to (B, b)$ and $g : (B, b) \to (C, c)$ are algebra homomorphisms, then $(g \circ f) \circ a = g \circ b \circ T (f) = c \circ T (g) \circ T (f) = c \circ T (g \circ f)$ , so their composite is one too.

3.1step 1.1step 2.1∎

Composition of these morphisms is the composition in $C$ ; step 1.1 supplies its identities and step 2.1 its closure, while associativity and the identity laws are inherited from $C$ .

Depends on

Algebra and algebra homomorphism for a monad

Used by

Eilenberg–Moore category of a monad Definition

Dependency tree · two levels

2 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

E. Riehl, Category Theory in Context, 2nd ed., Definition 5.2.4 (standard reference, not scraped)