PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-29

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A map from a disjoint union is smooth iff each restriction is smooth

Statement

Let $X = ⨆_{i \in I} M_{i}$ be a countable disjoint union of fixed-dimensional smooth manifolds with its canonical smooth structure, let $N$ be a smooth manifold, and let $F : X \to N$ be a map. Then $F$ is smooth if and only if each restriction

$F_{i} : = F \circ κ_{i} : M_{i} \to N$

to a summand is smooth.

Facts & Assumptions

Given: A countable disjoint union $X = ⨆_{i \in I} M_{i}$ with canonical injections $κ_{i} : M_{i} \to X$ , a smooth manifold $N$ , and a map $F : X \to N$ .

[F1]

The disjoint union $X$ is a smooth manifold whose smooth charts are the transported charts coming from the summands; in particular every point of $X$ lies in exactly one summand and around that point there are smooth charts coming from that summand (Countable disjoint unions of fixed-dimensional smooth manifolds are smooth manifolds).

[F2]

A map between smooth manifolds is smooth exactly when it is continuous and its coordinate representatives are smooth near each point ( $C^{r}$ and smooth maps between smooth manifolds).

[L1]

A map out of a disjoint union is continuous exactly when each restriction to a summand is continuous (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).

Proof

technique · direct

1.1givenF1F2L1choose

Assume $F$ is smooth, and fix $i \in I$ and $p \in M_{i}$ with charts as below. [given, F1, F2, L1, choose] By [F2] it is continuous, so [L1] makes each restriction $F_{i} = F \circ κ_{i}$ continuous. Fix $i \in I$ and $p \in M_{i}$ . Choose a smooth chart $(U, α)$ of $M_{i}$ at $p$ and a smooth chart $(V, β)$ of $N$ at $F_{i} (p)$ . The transported chart $(κ_{i} [U], α \circ κ_{i}^{- 1})$ is a smooth chart of $X$ at $κ_{i} (p)$ by [F1].

1.2F1F2L1choose

Conversely assume every restriction $F_{i}$ is smooth, and fix $x \in X$ with charts as below. [F1, F2, L1, choose] Then [F2] makes each $F_{i}$ continuous, so [L1] makes $F$ continuous. Let $x \in X$ . By [F1] there is a unique $i \in I$ and a unique point $p \in M_{i}$ with $x = κ_{i} (p)$ ; choose a smooth chart $(U, α)$ of $M_{i}$ at $p$ and a smooth chart $(V, β)$ of $N$ at $F (x)$ . The transported chart $(κ_{i} [U], α \circ κ_{i}^{- 1})$ is smooth on $X$ .

2.1F1F2step 1.1step 1.2

In the charts chosen in steps 1.1 and 1.2, the representative of the [F1, F2, step 1.1, step 1.2] restriction $F_{i}$ is $β \circ F_{i} \circ α^{- 1} = β \circ F \circ (α \circ κ_{i}^{- 1})^{- 1}$ . Under the hypothesis of step 1.1, the right-hand side is the representative of $F$ in a transported source chart, so [F2] makes it smooth and therefore every $F_{i}$ is smooth. Under the hypothesis of step 1.2, the same formula identifies the representative of $F$ with $β \circ F_{i} \circ α^{- 1}$ , which is smooth because $F_{i}$ is. Hence [F2] makes $F$ smooth at $x$ , and therefore smooth everywhere.

3.1step 2.1∎

Step 2.1 proves both directions of the equivalence.

Depends on

Used by

Restrictions, corestrictions, and products of smooth maps are smooth Proposition
Cartesian product makes bordism a graded ring Theorem
Disjoint union makes bordism classes abelian groups Theorem

Dependency tree · two levels

19 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

Rob van der Vorst, Introduction to differentiable manifolds, §2 (standard reference, not scraped)
Nigel Hitchin, Differentiable Manifolds, §2.4 (standard reference, not scraped)