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

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:

1. 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.
2. 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 $\bigcup_{i \in I} U_i = X$. If $f|_{U_i} : U_i \to Y$ is continuous for every $i \in I$, then $f$ is continuous.
3. Finite closed cover. Let $f : X \to Y$ be a function, let $n \ge 1$ 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.

• The finiteness in claim 3 is not removable. The witness is on the companion page: $\mathbb{R}$ 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 ($\mathbb{R}$ 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.