False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-17

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FALSE: Every functor with a left adjoint is monadic

Statement

False claim: every functor that has a left adjoint is monadic.

The underlying-set functor $U : T o p \to S e t$ is a counterexample.

Facts & Assumptions

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

[L1]

The discrete-topology functor is left adjoint to $U$ (Discrete topology, underlying set, and indiscrete topology form an adjoint triple).

[L2]

Every monadic functor is conservative (Every monadic functor is conservative).

[L3]

A continuous bijection is an isomorphism in $T o p$ exactly when its inverse is continuous (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

Refutation

technique · direct

1.1L1

By [L1], $U$ has a left adjoint.

1.2L3

On a two-element set $X$ , let $X_{d}$ have the discrete topology and $X_{i}$ the indiscrete topology. The identity function $q : X_{d} \to X_{i}$ is continuous and $U (q)$ is a bijection, but $q^{- 1} : X_{i} \to X_{d}$ is not continuous because a singleton is open in $X_{d}$ and not in $X_{i}$ .

2.1L2step 1.1step 1.2∎

Thus $U (q)$ is an isomorphism while $q$ is not, so $U$ is not conservative. By [L2] it is not monadic, even though step 1.1 gives it a left adjoint.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

24 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 5.3.1 and Theorem 5.5.1 (standard reference, not scraped)