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In this post I wrote down the definition of an abstract manifold. Let me repeat it here:

Definition. Let $M$ be metric space and $\left(V_{i}\right)_{i\in I}$ be an open cover of $M$ with open sets $U_i\subset\mathbb R^m$ and homeomorphisms $F_i: U_i\rightarrow V_i$. Then we say $M$ is an (abstract) $m$-dimensional $C^k$ manifold if for all two open sets $V_1$, $V_2\subset M$ with maps $F_1$ and $F_2$ the transition map $$F_{2}^{-1}\circ F_1: \quad F_1^{-1}(V_1 \cap V_2) \rightarrow F_{2}^{-1}(V_1\cap V_2)$$ is a $C^k$ diffeomorphism.

I realized that I have not fully captured yet the part with the transition maps. Why exactly is it $F_{2}^{-1}\circ F_1: F_1^{-1}(V_1 \cap V_2) \rightarrow F_{2}^{-1}(V_1\cap V_2)$ and not $F_{2}^{-1}\circ F_1: U_1 \rightarrow U_2$? After all, the domain of $F_1$ is $U_1$ and the codomain of $F_2^{-1}$ is $U_2$.

José Carlos Santos
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Hermi
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    If $p \in U_1$, why is $F_1(p)$ in the domain of $F_2^{-1}$? – Michael Albanese Aug 23 '21 at 17:49
  • Thanks. The domain of $F_{2}^{-1}$ is $V_2$, isn't it, and $F_1(p)\in V_2$ for $p\in U_1$. – Hermi Aug 23 '21 at 17:53
  • +1 (and +1 to @JoseCarlosSantos for a well written answer) -- while learning, you may find it more healthy to forget the notation and just think: if I've identified a neighborhood of a point with $\mathbb{R}^n$ in two different ways, then I have a map from $\mathbb{R}^n$ to itself and that map had better be $C^k$. – hunter Aug 23 '21 at 17:59
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    @Hermi: No, $F_1(p) \in V_1$ not necessarily $V_2$. – Michael Albanese Aug 23 '21 at 18:00

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The map $F_1$ is a map from $U_1$ to $V_1$. And $F_2^{\,-1}$ is a map from $V_2$ to $U_2$. But $F_2^{\,-1}\circ F_1$ is not defined on the whole $U_1$; $(F_2^{\,-1}\circ F_1)(x)$ makes sense only for those $x\in U_1$ such that $F_1(x)$ belongs to the domain of $F_2^{\,-2}$. And this happens exactly when $x\in F_1^{\,-1}(V_1\cap V_2)$, that is, when $x$ is such that $F_1(x)\in V_1\cap V_2$.

José Carlos Santos
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