If two charts are smoothly compatible with an atlas then they are smothly compatible with each other

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Let [ilmath]\mathcal{A}:\eq\{(U_\alpha,\varphi_\alpha)\}_{\alpha\in I} [/ilmath] be smooth atlas on some locally euclidean topological space, [ilmath](X,\mathcal{J})[/ilmath][Note 1] and suppose [ilmath](V,\psi)[/ilmath] and [ilmath](W,\sigma)[/ilmath] are charts, then[1]:

  • If we have both:
    1. [ilmath](V,\psi)[/ilmath] is smoothly compatible with the atlas [ilmath]\mathcal{A} [/ilmath] and
    2. [ilmath](W,\sigma)[/ilmath] is also smoothly compatible with the atlas [ilmath]\mathcal{A} [/ilmath]
  • then these two charts are themselves smoothly compatible with each other, i.e.:


We wish to show that [ilmath](V,\psi)[/ilmath] and [ilmath](W,\sigma)[/ilmath] are smoothly compatible charts, i.e. that the transition maps are smooth (in the usual [ilmath]\mathbb{R}^n[/ilmath] sense), that is to say that:

  • [ilmath](\sigma\circ\psi^{-1}):\underbrace{\psi(V\cap W)}_{\subseteq\mathbb{R}^m}\rightarrow\underbrace{\sigma(V\cap W)}_{\subseteq\mathbb{R}^n} [/ilmath][Note 2] is smooth in the real analysis sense
  • [ilmath](\psi\circ\sigma^{-1}):\sigma(V\cap W)\rightarrow\psi(V\cap W)[/ilmath] must also be smooth in the same sense (so we have a diffeomorphism)

Recall that to be considered smooth or [ilmath]C^\infty[/ilmath] they must be smooth at each point in the domain, we will show [ilmath]\forall p\in V\cap W[\sigma\circ\psi^{-1}\text{ is } [/ilmath][ilmath]\text{smooth} [/ilmath][ilmath]\text{ at }p][/ilmath], with this said we may now begin the proof:


  • If [ilmath]V\cap W[/ilmath] is empty then we're done, the statement is vacuously true
  • Otherwise:
    • Let [ilmath]p\in V\cap W[/ilmath] be given (as in this case there is a [ilmath]p[/ilmath] to speak of)
      • As [ilmath]\mathcal{A} [/ilmath] is a smooth atlas we see that [ilmath]X\subseteq\bigcup_{\alpha\in I}U_\alpha[/ilmath][Note 3]
        • By the implies-subset relation this is the same as:
          • [ilmath]\forall x\in X[x\in \bigcup_{\alpha\in I}U_\alpha][/ilmath], then
            • by the definition of union this is the same as (as in if and only if or [ilmath]\iff[/ilmath])
              • [ilmath]\forall x\in X[\exists \beta\in I[x\in U_\beta]][/ilmath] which we can re-write more neatly as:
              [ilmath]\forall x\in X\exists \beta\in I[x\in U_\beta][/ilmath]
      • Both [ilmath]V\subseteq X[/ilmath] and [ilmath]W\subseteq X[/ilmath] of course, so [ilmath]V\cap W\subseteq X[/ilmath][Note 4]
        • We see that [ilmath]p\in X[/ilmath] as a result (we already knew this but just to as formal as usual)
      • Thus: [ilmath]\exists \beta\in I[p\in U_\beta][/ilmath]
      • Define [ilmath]\beta\in I[/ilmath] to be such that [ilmath]p\in U_\beta[/ilmath] and [ilmath]\varphi_\beta[/ilmath] is a chart containing [ilmath]p[/ilmath]
        • Now [ilmath]p\in U_\beta\cap V\cap W[/ilmath]
        • By hypothesis the following four maps are smooth on their domains (by compatibility with the atlas):
          1. [ilmath](\psi\circ\varphi_\beta^{-1}):\varphi_\beta(V\cap U_\beta)\rightarrow\psi(V\cap U_\beta)[/ilmath] - which is one direction of the transition maps between [ilmath](V,\psi)[/ilmath] and [ilmath](U_\beta,\varphi_\beta)[/ilmath]
          2. [ilmath](\varphi_\beta\circ\psi^{-1}):\psi(V\cap U_\beta)\rightarrow\varphi_\beta(V\cap U_\beta)[/ilmath] - which is the other direction of the transition maps between [ilmath](V,\psi)[/ilmath] and [ilmath](U_\beta,\varphi_\beta)[/ilmath]
          3. [ilmath](\sigma\circ\varphi_\beta^{-1}):\varphi_\beta(W\cap U_\beta)\rightarrow\sigma(W\cap U_\beta)[/ilmath] - which is one direction of the transition maps between [ilmath](W,\sigma)[/ilmath] and [ilmath](U_\beta,\varphi_\beta)[/ilmath]
          4. [ilmath](\varphi_\beta\circ\sigma^{-1}):\sigma(W\cap U_\beta)\rightarrow\varphi_\beta(W\cap U_\beta)[/ilmath] - which is the other direction of the transition maps between [ilmath](W,\sigma)[/ilmath] and [ilmath](U_\beta,\varphi_\beta)[/ilmath]
Grade: A
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A bit more footwork is needed, the gist is:
  • [ilmath](\psi\circ\varphi_\beta^{-1})\big\vert_{\varphi_\beta(V\cap W\cap U_\beta)}\circ(\varphi_\beta\circ\sigma^{-1})\big\vert_{\sigma(V\cap W\cap U_\beta)} [/ilmath] - notice the restrictions in play - as this composition is not defined, but can be defined on a subset of the domain of each part!
    • This requires us to show that [ilmath]\left((\varphi_\beta\circ\sigma^{-1})\big\vert_{\sigma(V\cap W\cap U_\beta)}\right)(\sigma(V\cap W\cap U_\beta))\subseteq\varphi_\beta(V\cap W\cap U_\beta)[/ilmath] for the composition of the restrictions to be defined
  • Then we say:
    • [ilmath](\psi\circ\varphi_\beta^{-1})\big\vert_{\varphi_\beta(V\cap W\cap U_\beta)}\circ(\varphi_\beta\circ\sigma^{-1})\big\vert_{\sigma(V\cap W\cap U_\beta)} [/ilmath]
      [ilmath]\eq(\psi\circ\sigma^{-1})\big\vert_{\sigma(V\cap W\cap U_\beta)} [/ilmath]
  • Then as [ilmath](\varphi_\beta\circ\sigma^{-1})[/ilmath] and [ilmath](\psi\circ\varphi_\beta^{-1})[/ilmath] are smooth (by hypothesis)
    • so are their restrictions
      TODO: Are they though?
    • Thus so is [ilmath](\psi\circ\sigma^{-1})\big\vert_{\sigma(V\cap W\cap U_\beta)} [/ilmath] - specifically it is smooth at [ilmath]\sigma(p)[/ilmath]
      • If a map is smooth at a point in its domain then any extension of that map is smooth at that point
        TODO: Prove this too!
        • Thus we see [ilmath](\psi\circ\sigma^{-1})[/ilmath] is smooth at [ilmath]p[/ilmath]
  • Since [ilmath]p[/ilmath] was arbitrary we see the transition maps are smooth everywhere.

The other transition map is just the composition of the remaining 2 smooth transition maps we get by hypothesis and is basically the same work.

  • Time/date: Alec (talk) 13:38, 1 April 2017 (UTC)


  1. An Introduction to Manifolds - Loring W. Tu


  1. This page's statement is used to build a maximal smooth atlas and thus a smooth manifold from a topological manifold. We may speak of a (smooth) atlas on a space that is simply locally euclidean, it need not be a full on topological manifold so we relax that constraint
  2. Notice that we speak of [ilmath]\mathbb{R}^m[/ilmath] and [ilmath]\mathbb{R}^n[/ilmath], this is because the dimension of connected components may vary from component to component, in this case though [ilmath]V\cap W[/ilmath] would be empty of course.
  3. We actually have [ilmath]X\eq\bigcup_{\alpha\in I}U_\alpha[/ilmath] as the [ilmath]U_\alpha[/ilmath] are subsets of [ilmath]X[/ilmath] (in fact open sets so [ilmath]\in\mathcal{J} [/ilmath]) so we have [ilmath]\bigucp_{\alpha\in I}U_\alpha\subseteq X[/ilmath] automatically, we use [ilmath]X\subseteq\bigcup_{\alpha\in I}U_\alpha[/ilmath] for the atlas property to use the implies-subset relation to emphasise what we're doing. As we have just shown this is logically the same as equality between [ilmath]X[/ilmath] and the union
  4. TODO: What claim have I used here? The intersection of sets is a subset of each set might be related (or indeed good enough as then [ilmath]V\cap W\subseteq V\subseteq X[/ilmath]!)