CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11

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A functor need not preserve monomorphisms

Statement refuted

The assertion that every functor preserves monomorphisms is false.

Facts & Assumptions

Given: The walking-arrow category $\mathbf2=(0\xrightarrow{m}1)$ and $\mathbf{Set}$ .

[L1]

A functor preserves identities and composition (Covariant functor, identity functor, composite functor, and contravariant functor).

[L2]

A monomorphism is left-cancellative (Monomorphism and epimorphism by left and right cancellation).

[L3]

Sets and functions form $\mathbf{Set}$ (Sets and functions form the large locally small category $\mathbf{Set}$ ).

Counterexample

technique · direct

1.1

The only morphism of $\mathbf2$ with codomain $0$ is $1_0$ . Thus any parallel $r,s$ with $mr=ms$ must both equal $1_0$ , so $m$ is monic by [L2].

1.2

Define a functor $F:\mathbf2\to\mathbf{Set}$ by $F(0)=\{0,1\}$ , $F(1)=\{*\}$ , and $F(m)$ the constant function. This assignment respects all possible identities and composites, so it is a functor.

L1L3

2.1

Let $r,s:\{*\}\to\{0,1\}$ select $0$ and $1$ , respectively. Then $r\ne s$ but $F(m)r=F(m)s$ , so $F(m)$ is not monic by [L2].

step 1.2L2

3.1

The monomorphism $m$ of step 1.1 is sent by the functor $F$ to the nonmonomorphism of step 2.1. This is the required counterexample.

step 1.1step 2.1∎

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 19 results over 8 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.