TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A map out of a disjoint union is continuous iff each of its restrictions is; the canonical injections are open and closed embeddings; and each summand is clopen in the union

Statement

Let $(X_i, \mathcal{T}_i)_{i \in I}$ be topological spaces and let $S := \bigsqcup_{i \in I} X_i$ carry the disjoint union topology, with canonical injections $\kappa_j$ (The disjoint union (coproduct) $\bigsqcup_i X_i$ with the final topology of the canonical injections: a set is open exactly when each of its traces is). Then:

Characteristic property. For every space $W$ and every function $k : S \to W$ , $k \text{ is continuous } \iff k \circ \kappa_i \text{ is continuous for every } i \in I ,$ and every family of continuous maps $k_i : X_i \to W$ arises from exactly one such $k$ , namely $k(x,i) := k_i(x)$ .
The injections are continuous, open and closed and injective (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological, Injection, surjection, bijection); consequently each $\kappa_j$ is an embedding, and the subspace topology on $\kappa_j[X_j]$ is the image of $\mathcal{T}_j$ under $\kappa_j$ (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).
Each summand is clopen. $\kappa_j[X_j] = X_j \times \{j\}$ is both open and closed in $S$ , and the sets $\kappa_j[X_j]$ , $j \in I$ , are pairwise disjoint with union $S$ .

Facts & Assumptions

Given: Topological spaces $(X_i,\mathcal{T}_i)_{i \in I}$ , the set $S = \bigsqcup_i X_i$ with the disjoint union topology, the injections $\kappa_j(x) = (x,j)$ , an index $j \in I$ , a space $W$ and a function $k : S \to W$ .

[A1]

$S = \bigcup_i (X_i \times \{i\})$ ; each $\kappa_i$ is injective; the sets $X_i \times \{i\}$ are pairwise disjoint with union $S$ ; and $U \subseteq S$ is open exactly when $\kappa_i^{-1}[U]$ is open in $X_i$ for every $i$ , closed exactly when every $\kappa_i^{-1}[U]$ is closed (The disjoint union (coproduct) $\bigsqcup_i X_i$ with the final topology of the canonical injections: a set is open exactly when each of its traces is, Injection, surjection, bijection).

[A2]

The disjoint union topology is the final topology of the family $(\kappa_i)_{i \in I}$ (The disjoint union (coproduct) $\bigsqcup_i X_i$ with the final topology of the canonical injections: a set is open exactly when each of its traces is).

[L1]

For a final topology of a family $(g_i)$ : each $g_i$ is continuous, and a map $k$ out of the space is continuous exactly when every $k \circ g_i$ is (Characteristic properties: a map into a space with the initial topology is continuous iff every composite with the defining family is, a map out of a space with the final topology is continuous iff every composite with the defining family is, and the two topologies are respectively the coarsest and the finest making that family continuous, claims 3 and 4; Continuity of a map of topological spaces at a point and globally).

[L2]

$f$ is an open map when images of open sets are open, a closed map when images of closed sets are closed, and an embedding when it is injective and its corestriction to its image, with the subspace topology, is a homeomorphism (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological, 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).

[L3]

A map into a subspace is continuous exactly when its composite with the inclusion 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); a continuous bijection is a homeomorphism exactly when it is open (A continuous bijection is a homeomorphism iff it is open iff it is closed, and homeomorphy is an equivalence relation on spaces, claim 1).

[L4]

A topology contains the empty set and the whole space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

Proof

technique · direct

1.1

By [A2] and [L1] the injections are continuous and claim 1's equivalence holds.

A2L1

1.2

A family of functions $k_i : X_i \to W$ determines exactly one $k : S \to W$ with $k \circ \kappa_i = k_i$ for every $i$ : every element of $S$ is $(x,i)$ for exactly one pair by [A1], so $k(x,i) := k_i(x)$ is a well defined function, and any $k'$ with $k' \circ \kappa_i = k_i$ agrees with it at every $(x,i)$ .

1.3

Let $V \subseteq X_j$ and compute the traces of $\kappa_j[V] = V \times \{j\}$ : for $i = j$ the trace is $V$ , and for $i \ne j$ it is $\varnothing$ , since $(x,i) \in V \times \{j\}$ forces $i = j$ .

2.1

If $V$ is open in $X_j$ , then by step 1.3 all traces of $\kappa_j[V]$ are open, $\varnothing$ being open by [L4]; so $\kappa_j[V]$ is open in $S$ by [A1] and $\kappa_j$ is an open map.

step 1.3A1L2L4

2.2

If $F$ is closed in $X_j$ , then by step 1.3 the traces of $\kappa_j[F]$ are $F$ and $\varnothing$ , both closed, $\varnothing$ being closed by [L4]; so $\kappa_j[F]$ is closed in $S$ by [A1] and $\kappa_j$ is a closed map.

step 1.3A1L2L4

2.3

The corestriction $\kappa_j^0 : X_j \to \kappa_j[X_j]$ is a bijection, being injective by [A1] and surjective onto its image, and it is continuous by [L3], since $\kappa_j$ is continuous by step 1.1.

step 1.1A1L3

3.1

Taking $V := X_j$ in step 2.1 and $F := X_j$ in step 2.2 shows that $\kappa_j[X_j]$ is open and closed in $S$ ; with the disjointness and the covering property of [A1] this is claim 3.

step 2.1step 2.2A1L4

3.2

$\kappa_j^0$ is an open map into the subspace $\kappa_j[X_j]$ : for $V$ open in $X_j$ the set $\kappa_j[V]$ is open in $S$ by step 2.1 and is contained in $\kappa_j[X_j]$ , so it equals its own trace on $\kappa_j[X_j]$ and is open there.

step 2.1L2

4.1

By steps 2.3 and 3.2 with [L3] the map $\kappa_j^0$ is a homeomorphism onto the subspace $\kappa_j[X_j]$ , so $\kappa_j$ is an embedding and the subspace topology on $\kappa_j[X_j]$ is the image of $\mathcal{T}_j$ ; with steps 1.1, 2.1 and 2.2 this is claim 2.

step 1.1step 2.1step 2.2step 2.3step 3.2L2L3

5.1

Step 1.1 and step 1.2 give claim 1, step 4.1 gives claim 2 and step 3.1 gives claim 3.

step 1.1step 1.2step 3.1step 4.1∎

Remarks

The coproduct is where "define a map piecewise" becomes a theorem. Claim 1 says that specifying a continuous map on each summand separately, with no compatibility condition whatever, specifies a continuous map on the union. The absence of a compatibility condition is exactly what the disjointness buys; the pasting lemma (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous) is the corresponding statement for covers that do overlap, and it needs the pieces to agree.
Being open and closed is unusual, and it is what separates the summands. A continuous map out of $S$ can be constant on one summand and wild on another, so no summand is topologically attached to any other. This is the reason the disjoint union appears in the construction of an adjunction space: the gluing is put in afterwards, by a quotient, and the coproduct contributes no gluing of its own.
Nothing here needs the index set to be small. Claims 1 to 3 hold for an arbitrary index set and no choice principle is used, the maps in every step being given by explicit formulas.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 35 results over 12 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

Disjoint union (topology) (Wikipedia) (standard reference, not scraped)
Coproduct (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §22 (standard reference, not scraped)