Smoothness is a chart-independent statement about one map
Objective: Why is the single formula θ↦2θ on (−π,π) not by itself a complete global proof?
A manifold has many valid coordinate systems. A formula may look different after changing coordinates, so differential topology cannot define smoothness by favoring one chart. Instead, every compatible source and target chart must turn the map into an ordinary smooth Euclidean function wherever their domains meet. Before calculating, name the source manifold, target manifold, dimensions, charts, map, point, tangent spaces, and every regularity or compactness hypothesis. The exact conclusion to understand is: It is enough to verify smoothness in one smooth atlas on M and one smooth atlas on N; every other compatible chart gives a smooth coordinate expression automatically.
The new object means: Let M and N be smooth manifolds. A map f:M→N is smooth when for every chart (U,φ) of M and every chart (V,ψ) of N, the coordinate expression ψ∘f∘φ^{-1} is a smooth map between open subsets of Euclidean spaces on φ(U∩f^{-1}(V)). The symbol ∘ means composition: apply the rightmost map first. Read every derivative as a linear map between named tangent spaces and every geometric picture as evidence for a statement that still needs a coordinate-independent proof.
Rebuild the worked problem “Prove that f:S¹→S¹ given in complex notation by f(z)=z² is smooth, using the angle charts away from one cut point and checking what happens on chart overlaps.” from the definitions. Verify dimensions and hypotheses before invoking a theorem, and keep this failure boundary visible: A continuous map need not be smooth, and a formula smooth in one arbitrary coordinate parameter is not enough unless that parameter belongs to the declared smooth atlas. At a chart boundary, one must move to an overlapping chart rather than differentiate outside a chart domain.
It is enough to verify smoothness in one smooth atlas on M and one smooth atlas on N; every other compatible chart gives a smooth coordinate expression automatically.
Take a verified coordinate expression g=ψ∘f∘φ^{-1}. In new charts φ̃ and ψ̃, insert identity maps to write ψ̃∘f∘φ̃^{-1}=(ψ̃∘ψ^{-1})∘g∘(φ∘φ̃^{-1}).
The two outside maps are transition maps between compatible smooth charts, so both are smooth on their named overlap domains.
A composition of smooth Euclidean maps is smooth. Therefore the new coordinate expression is smooth, proving the definition does not depend on the chosen atlas charts.
Prove that f:S¹→S¹ given in complex notation by f(z)=z² is smooth, using the angle charts away from one cut point and checking what happens on chart overlaps.
- On an angle chart write z=e^{iθ}. Wherever the target angle branch is fixed, f(e^{iθ})=e^{i2θ}, so the coordinate expression is θ↦2θ followed, if needed, by adding an integer multiple of 2π to remain in the chosen target interval.
- The map θ↦2θ+2πm is an ordinary smooth real function on every connected overlap piece, because m is constant on that piece.
- Angle-chart transition maps also add integer multiples of 2π. Composing with them preserves smoothness, so all chart expressions are smooth and f is a smooth circle map.
Result: Every local angle expression is affine θ↦2θ+2πm on its overlap component; hence z↦z² is smooth on all of S¹.