TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-29 (claude-fable-5)

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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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A continuous image of a connected space is connected, and connectedness is a topological property

Statement

Let $X$ and $Y$ be topological spaces and let $f : X \to Y$ be continuous (Continuity of a map of topological spaces at a point and globally). Subsets carry the subspace topology (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace). Then:

Images. If $A \subseteq X$ is a connected subset of $X$ (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets) then $f [A]$ is a connected subset of $Y$ . In particular, if $X$ is connected then $f [X]$ is connected, and if $f$ is moreover surjective then $Y$ is connected.
Topological invariance. If $h : X \to Y$ is a homeomorphism (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological) then $X$ is connected if and only if $Y$ is. So connectedness is a topological property.

Nothing is assumed about $f$ beyond continuity: it need not be injective, open, closed or surjective. Note the direction — a continuous image of a connected space is connected, while a continuous preimage need not be, since a constant map from a disconnected space is continuous.

Facts & Assumptions

Given: Topological spaces $X$ and $Y$ , a continuous map $f : X \to Y$ , and a subset $A \subseteq X$ .

[A1]

A subset $S$ of a space is connected exactly when every continuous map $S \to 2$ is constant, $2$ being the two-point discrete space (For a topological space the following agree: no separation exists, the only clopen subsets are $\emptyset$ and $X$ , and every continuous map to the two-point discrete space is constant, claim 2, Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets).

[A2]

A restriction $f ∣_{A} : A \to Y$ of a continuous map to a subspace is continuous (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

[A3]

Characteristic property of a map into a subspace: for $S \subseteq Y$ with inclusion $ι : S \to Y$ and a function $g : Z \to S$ , the map $g$ is continuous exactly when $ι \circ g$ is (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

[A4]

A composite of continuous maps is continuous (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous, claim 1).

[A5]

A homeomorphism is a continuous bijection whose inverse is continuous, and a bijection is surjective (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

Proof

technique · direct

1.1

Write $g : A \to f [A]$ for the map $g (a) = f (a)$ , which is well defined because $f (a) \in f [A]$ for $a \in A$ , and is surjective by the definition of the image $f [A] = {f (a) : a \in A}$ .

given

1.2

The composite of $g$ with the inclusion $ι : f [A] \to Y$ is the restriction $f ∣_{A}$ , which is continuous by [A2]; so $g$ is continuous by [A3] applied with $Z = A$ and $S = f [A]$ .

A2A3

2.1

Assume $A$ is a connected subset of $X$ and let $χ : f [A] \to 2$ be continuous. Then $χ \circ g : A \to 2$ is continuous by step 1.2 and [A4], hence constant by [A1] applied to $A$ .

step 1.2A1A4

3.1

Since $g$ is surjective by step 1.1, every pair of points of $f [A]$ is of the form $g (a_{1}), g (a_{2})$ , and $χ (g (a_{1})) = χ (g (a_{2}))$ by step 2.1; so $χ$ is constant.

step 1.1step 2.1

4.1

As $χ$ was an arbitrary continuous map $f [A] \to 2$ , [A1] gives that $f [A]$ is a connected subset of $Y$ . Taking $A = X$ gives that $f [X]$ is connected when $X$ is, and if $f$ is surjective then $f [X] = Y$ , so $Y$ is connected. This is claim 1.

step 3.1A1

5.1

For claim 2 let $h : X \to Y$ be a homeomorphism. If $X$ is connected then $Y = h [X]$ is connected by step 4.1, since $h$ is continuous and surjective by [A5]; and if $Y$ is connected then $X = h^{- 1} [Y]$ is connected by step 4.1 applied to the continuous surjection $h^{- 1}$ , again by [A5]. So connectedness is preserved in both directions by a homeomorphism.

step 4.1A5∎

Remarks

Why the corestriction is the only technical point. Claim 1 is about $f [A]$ as a space, so the map that must be shown continuous is the one landing in $f [A]$ , not the one landing in $Y$ . The characteristic property of a subspace is exactly the tool that upgrades the second to the first, and it is the reason the proof needs no hypothesis on $f$ at all.
The hypothesis cannot be moved to the target. If $f [X]$ is connected nothing follows about $X$ : the constant map from any space whatever has a one-point image, which is connected. So claim 1 is a one-way implication and is used only in that direction below.
What the theorem buys immediately. Any property preserved by continuous images can be checked on a convenient model. That is the whole mechanism behind the intermediate value theorem in the next item, and behind the connectedness of every path-connected space later on this page: both work by pushing a connected interval forward along a continuous map.

Depends on

Used by

Dependency tree · two levels

15 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

Connected space (Wikipedia) (standard reference, not scraped)
The Stacks Project, Section 5.7: Connected components (standard reference, not scraped)