LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous

Statement

Let $X$ , $Y$ and $Z$ be topological spaces, with subspaces carrying 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:

Composites. If $f : X \to Y$ and $g : Y \to Z$ are continuous (Continuity of a map of topological spaces at a point and globally) then $g \circ f : X \to Z$ is continuous.
Open cover. Let $f : X \to Y$ be a function and let ${U_{i} : i \in I}$ be a family of open subsets of $X$ with $⋃_{i \in I} U_{i} = X$ . If $f ∣_{U_{i}} : U_{i} \to Y$ is continuous for every $i \in I$ , then $f$ is continuous.
Finite closed cover. Let $f : X \to Y$ be a function, let $n \geq 1$ and let $F_{1}, \dots, F_{n}$ be closed subsets of $X$ with $F_{1} \cup \dots \cup F_{n} = X$ . If $f ∣_{F_{k}} : F_{k} \to Y$ is continuous for every $k$ , then $f$ is continuous.

The converses of claims 2 and 3 hold with no hypothesis on the cover at all: every restriction 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). The finiteness in claim 3 is not removable; see the remarks.

Facts & Assumptions

Given: Topological spaces $X$ , $Y$ , $Z$ ; functions $f : X \to Y$ and $g : Y \to Z$ ; a family ${U_{i} : i \in I}$ of open subsets of $X$ covering $X$ ; a natural $n \geq 1$ and closed subsets $F_{1}, \dots, F_{n}$ of $X$ covering $X$ . For $S \subseteq X$ and $W \subseteq Y$ one has $(f ∣_{S})^{- 1} [W] = f^{- 1} [W] \cap S$ , and $(g \circ f)^{- 1} [W^{'}] = f^{- 1} [g^{- 1} [W^{'}]]$ for $W^{'} \subseteq Z$ .

[A1]

$f$ is continuous if and only if preimages of open sets are open, if and only if preimages of closed sets are closed (For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f (\overline{A}) \subseteq \overline{f (A)}$ , clauses (b) and (c)).

[A2]

The subspace topology on $S \subseteq X$ has as its open sets the traces $U \cap S$ with $U$ open in $X$ , and as its closed sets the traces $F \cap S$ with $F$ closed in $X$ ; if $S$ is open in $X$ then every set open in $S$ is open in $X$ , and if $S$ is closed in $X$ then every set closed in $S$ is closed in $X$ (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).

[L1]

A topology is closed under arbitrary unions of open sets (T2), and its closed sets are closed under finite unions (C3) (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

Proof

technique · direct

1.1givenA1

Claim 1: let $W \subseteq Z$ be open; then $g^{- 1} [W]$ is open in $Y$ and hence $f^{- 1} [g^{- 1} [W]]$ is open in $X$ , and this set is $(g \circ f)^{- 1} [W]$ ; so $g \circ f$ is continuous.

1.2givenA1A2

Let $V \subseteq Y$ be open. For each $i \in I$ the set $f^{- 1} [V] \cap U_{i} = (f ∣_{U_{i}})^{- 1} [V]$ is open in the subspace $U_{i}$ , because $f ∣_{U_{i}}$ is continuous; and $U_{i}$ is open in $X$ , so this set is open in $X$ .

1.3givenA1A2

Let $F \subseteq Y$ be closed. For each $k \leq n$ the set $f^{- 1} [F] \cap F_{k} = (f ∣_{F_{k}})^{- 1} [F]$ is closed in the subspace $F_{k}$ , because $f ∣_{F_{k}}$ is continuous; and $F_{k}$ is closed in $X$ , so this set is closed in $X$ .

2.1step 1.2givenA1L1

Since the $U_{i}$ cover $X$ , $f^{- 1} [V] = ⋃_{i \in I} (f^{- 1} [V] \cap U_{i})$ , a union of sets open in $X$ by step 1.2, hence open in $X$ by (T2). As $V$ was an arbitrary open subset of $Y$ , $f$ is continuous, which is claim 2.

2.2step 1.3givenA1L1

Since the $F_{k}$ cover $X$ , $f^{- 1} [F] = ⋃_{k = 1}^{n} (f^{- 1} [F] \cap F_{k})$ , a union of finitely many sets closed in $X$ by step 1.3, hence closed in $X$ by (C3) iterated, the union being over $n \geq 1$ sets. As $F$ was an arbitrary closed subset of $Y$ , $f$ is continuous, which is claim 3.

3.1step 1.1step 2.1step 2.2∎

Claims 1, 2 and 3 are established by step 1.1, step 2.1 and step 2.2 respectively.

Remarks

The finiteness in claim 3 is not removable. The witness is on the companion page: $R$ with its usual topology is covered by its closed singletons, every restriction of the indicator function of ${0}$ to a singleton is continuous, and that function is not continuous ( $R$ covered by its closed singletons: every restriction of the indicator of ${0}$ is continuous and the map is not, so the closed pasting lemma needs finiteness ↗). No corresponding restriction is needed in claim 2.
Where each hypothesis is spent. Claim 2 uses openness of the cover members only to pass from "open in $U_{i}$ " to "open in $X$ ", and it allows an arbitrary index set because arbitrary unions of open sets are open. Claim 3 uses closedness of the cover members for the corresponding passage, and it must restrict to finitely many because only finite unions of closed sets are closed. The two asymmetries of the topology axioms are visible in the two statements, one each.
The usual two-piece form. Claim 3 with $n = 2$ is the pasting lemma as it is normally quoted: if $X = F_{1} \cup F_{2}$ with both pieces closed and $f_{1} : F_{1} \to Y$ , $f_{2} : F_{2} \to Y$ are continuous and agree on $F_{1} \cap F_{2}$ , then the combined function is well defined and continuous. Well definedness is the agreement hypothesis and is not a topological matter; continuity is claim 3.
Continuity is a local property, and claim 2 is the precise sense. A function continuous in a neighbourhood of each point is continuous, because the interiors of those neighbourhoods form an open cover. No such statement holds for uniform notions, which is why nothing here is called uniform.

Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous

Statement

Facts & Assumptions

Proof

Remarks

Depends on

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Sources