False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-16

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Every functor with a left adjoint also has a right adjoint

Statement

If a functor has a left adjoint, then it also has a right adjoint.

Facts & Assumptions

Given: The underlying-set functor $U : G r p \to S e t$ .

[F1]

The free-group functor is a left adjoint of $U$ (The free-group functor is left adjoint to the underlying-set functor).

[F2]

Every left adjoint preserves colimits, including initial objects (Left adjoints preserve every colimit that exists).

[F3]

An initial object has exactly one morphism to every object (Initial object, terminal object, and zero object).

Refutation

technique · contradiction

1.1F1F2assume-contra

By [F1], $U$ has a left adjoint. Suppose, as the claim predicts, that $U$ also has a right adjoint. Then $U$ itself is a left adjoint and preserves initial objects by [F2].

1.2F3algebra

The trivial group is initial in $G r p$ , since there is exactly one homomorphism from it to every group. Its underlying set is a singleton.

2.1step 1.1step 1.2F3discharge-contradiction∎

The empty set, not a singleton, is initial in $S e t$ , so step 1.2 contradicts the preservation conclusion of step 1.1. Hence $U$ has no right adjoint and the statement is false.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

11 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., Example 4.1.10 (standard reference, not scraped)