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A map from a disjoint union is smooth iff each restriction is smooth
Statement
Let be a countable disjoint union of fixed-dimensional smooth manifolds with its canonical smooth structure, let be a smooth manifold, and let be a map. Then is smooth if and only if each restriction
to a summand is smooth.
Facts & Assumptions
Given: A countable disjoint union with canonical injections , a smooth manifold , and a map .
The disjoint union is a smooth manifold whose smooth charts are the transported charts coming from the summands; in particular every point of 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).
A map between smooth manifolds is smooth exactly when it is continuous and its coordinate representatives are smooth near each point ( and smooth maps between smooth manifolds).
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
Assume is smooth, and fix and with charts as below. [given, F1, F2, L1, choose] By [F2] it is continuous, so [L1] makes each restriction continuous. Fix and . Choose a smooth chart of at and a smooth chart of at . The transported chart is a smooth chart of at by [F1].
Conversely assume every restriction is smooth, and fix with charts as below. [F1, F2, L1, choose] Then [F2] makes each continuous, so [L1] makes continuous. Let . By [F1] there is a unique and a unique point with ; choose a smooth chart of at and a smooth chart of at . The transported chart is smooth on .
In the charts chosen in steps 1.1 and 1.2, the representative of the [F1, F2, step 1.1, step 1.2] restriction is . Under the hypothesis of step 1.1, the right-hand side is the representative of in a transported source chart, so [F2] makes it smooth and therefore every is smooth. Under the hypothesis of step 1.2, the same formula identifies the representative of with , which is smooth because is. Hence [F2] makes smooth at , and therefore smooth everywhere.
Step 2.1 proves both directions of the equivalence.
Depends on
- Countable disjoint unions of fixed-dimensional smooth manifolds are smooth manifolds
- $C^r$ and smooth maps between smooth manifolds
- 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
Used by
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)