TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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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

1.1

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 $T o p (D X, Y) ≅ S e t (X, U Y)$ .

F1F2F3F4

1.2

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 $S e t (U Y, X) ≅ T o p (Y, I X)$ .

F1F2F3F4

2.1

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

step 1.1step 1.2

3.1

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

step 2.1L1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 57 results over 20 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.

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)