TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16

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Discrete topology, underlying set, and indiscrete topology form an adjoint triple

Statement

Let $D, I : S e t \to T o p$ equip a set with the discrete and indiscrete topology, respectively, and let $U : T o p \to S e t$ forget the topology. Then

$D ⊣ U ⊣ I .$

Facts & Assumptions

Given: A set $X$ and a topological space $Y$ .

[F1]

The discrete topology on a set is its power set, and the indiscrete topology is ${\emptyset, X}$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

[F2]

A map $f$ is continuous when it is continuous at every point, and $f$ is continuous at $x$ exactly when for every open $V \subseteq Y$ with $f (x) \in V$ there is an open $U \subseteq X$ with $x \in U$ and $f [U] \subseteq V$ (Continuity of a map of topological spaces at a point and globally).

[F3]

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

[F4]

Topological spaces and continuous maps form the locally small category $T o p$ (Topological spaces and continuous maps form the large locally small category $T o p$ ).

[L1]

An adjoint triple consists of adjunctions on both sides of its middle functor (Adjoint triple $L ⊣ M ⊣ R$ ).

Proof

technique · direct

1.1F1F2F3F4

Every function $f : X \to U Y$ is continuous as a map $D (X) \to Y$ : given $x \in X$ and an open $V ∋ f (x)$ in $Y$ , the singleton $U = {x}$ is open in the discrete topology by [F1] and satisfies $f [U] \subseteq V$ , so [F2] gives continuity at $x$ and hence continuity. Thus the identity-on-functions correspondence gives $T o p (D X, Y) ≅ S e t (X, U Y)$ .

1.2F1F2F3F4

Every function $g : U Y \to X$ is continuous as a map $Y \to I (X)$ : by [F1] the only open sets of $I (X)$ are $\emptyset$ and $X$ , so an open $V$ containing $g (y)$ must be $X$ , and $U = Y$ is open with $g [U] \subseteq X$ ; [F2] again gives continuity. Thus $S e t (U Y, X) ≅ T o p (Y, I X)$ .

2.1step 1.1step 1.2

Both correspondences are natural because precomposition and postcomposition leave the underlying function unchanged. Therefore $D ⊣ U$ and $U ⊣ I$ .

3.1step 2.1L1∎

By [L1] these adjunctions form the displayed triple. The same proof includes the empty set and singleton, whose discrete and indiscrete topologies may coincide.

Depends on

Used by

FALSE: Every functor with a left adjoint is monadic False statement

Dependency tree · two levels

22 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 4.1.6 (standard reference, not scraped)
Tom Leinster, Basic Category Theory, Example 2.1.5 (standard reference, not scraped)