CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck 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 $2 = (0 \overset{m}{\to} 1)$ and $S e t$ .

[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 $S e t$ (Sets and functions form the large locally small category $S e t$ ).

Counterexample

technique · direct

1.1

The only morphism of $2$ with codomain $0$ is $1_{0}$ . Thus any parallel $r, s$ with $m r = m s$ must both equal $1_{0}$ , so $m$ is monic by [L2].

1.2

Define a functor $F : 2 \to S e t$ 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 $r \neq s$ 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 · 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.