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

Statement

Let X=iIMi be a countable disjoint union of fixed-dimensional smooth manifolds with its canonical smooth structure, let N be a smooth manifold, and let F:XN be a map. Then F is smooth if and only if each restriction

Fi:=Fκi:MiN

to a summand is smooth.

Facts & Assumptions

Given: A countable disjoint union X=iIMi with canonical injections κi:MiX, a smooth manifold N, and a map F:XN.

[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 (Cr and smooth maps between smooth manifolds).

Proof

technique · direct
1.1

Assume F is smooth, and fix iI and pMi with charts as below. [given, F1, F2, L1, choose] By [F2] it is continuous, so [L1] makes each restriction Fi=Fκi continuous. Fix iI and pMi. Choose a smooth chart (U,α) of Mi at p and a smooth chart (V,β) of N at Fi(p). The transported chart (κi[U],ακi1) is a smooth chart of X at κi(p) by [F1].

givenF1F2L1choose
1.2

Conversely assume every restriction Fi is smooth, and fix xX with charts as below. [F1, F2, L1, choose] Then [F2] makes each Fi continuous, so [L1] makes F continuous. Let xX. By [F1] there is a unique iI and a unique point pMi with x=κi(p); choose a smooth chart (U,α) of Mi at p and a smooth chart (V,β) of N at F(x). The transported chart (κi[U],ακi1) is smooth on X.

F1F2L1choose
2.1

In the charts chosen in steps 1.1 and 1.2, the representative of the [F1, F2, step 1.1, step 1.2] restriction Fi is βFiα1=βF(ακi1)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 Fi is smooth. Under the hypothesis of step 1.2, the same formula identifies the representative of F with βFiα1, which is smooth because Fi is. Hence [F2] makes F smooth at x, and therefore smooth everywhere.

F1F2step 1.1step 1.2
3.1

Step 2.1 proves both directions of the equivalence.

step 2.1

Depends on

Used by

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