CorollaryStatement: AI-generatedProof: AI-generatedprecheck passaudited 2026-09-01

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In a cartesian closed category, any initial object is strict

Statement

Let $C$ be a cartesian closed category and let $0$ be an initial object. Then $0$ is strict: every morphism $f : A \to 0$ is an isomorphism.

Facts & Assumptions

Given: A cartesian closed category $C$ , an initial object $0$ , an object $A$ , and a morphism $f : A \to 0$ .

[L1]

In a cartesian closed category, the functor $A \times -$ is a left adjoint for every $A$ (Cartesian closed category).

[L2]

Left adjoints preserve colimits, hence preserve initial objects (Left adjoints preserve every colimit that exists).

[L3]

An initial object has exactly one endomorphism and exactly one map into any target (Initial object, terminal object, and zero object).

Proof

technique · direct

1.1givenL1L2

By [L1] and [L2], the functor $A \times -$ preserves the initial object, so $A \times 0$ is initial. Hence there is an isomorphism $e : A \times 0 \to 0$ .

2.1step 1.1givenalgebra

The pair $(1_{A}, f)$ induces a morphism $⟨ 1_{A}, f ⟩ : A \to A \times 0$ . Let $g : = π_{1} \circ e^{- 1} : 0 \to A$ , where $π_{1} : A \times 0 \to A$ is the first projection. Then $g \circ f = π_{1} \circ e^{- 1} \circ e \circ ⟨ 1_{A}, f ⟩ = π_{1} \circ ⟨ 1_{A}, f ⟩ = 1_{A}$ .

3.1step 2.1L3

The composite $f \circ g : 0 \to 0$ is the unique endomorphism of the initial object, so by [L3] it equals $1_{0}$ . Thus $g$ is a two-sided inverse to $f$ , and $f$ is an isomorphism.

4.1step 3.1∎

Therefore the initial object is strict.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 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

Emily Riehl, Category Theory in Context, 2nd ed., Theorem 4.2.1 (standard reference, not scraped)