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Definition: $F$ is a smooth map if for every $p \in M$, there exists smooth charts $(U,\phi)$ containing $p$ and $(V,\psi)$ containing $F(p)$ such that $F(U) \subseteq V$ and the composite map $\psi \circ F \ \circ \phi^{-1}$ is smooth from $\phi(U)$ to $\psi(V)$.

So, I'm trying to come up with examples of smooth maps on smooth manifolds and I'm running into a small issue. With the map $\psi \circ F \ \circ\phi^{-1} :\phi(U) \rightarrow \psi(V). $ For example, I tried $F: \mathbb R^{2} \rightarrow \mathbb R $ with $F(x,y) = d$ to be the constant function, then I'm trying to figure out what $\phi(U)$ and $\psi(V)$ should be. Also, would $\phi$ be a coordinate map from $\mathbb R^{2} \rightarrow \mathbb R$, and how would I similarly define $\psi$.

I'm finding the definition of smooth map to be a little confusing. My main problem is trying to come up with examples to make sense of how the charts work as well as the composition map $\psi\circ F \ \circ\phi^{-1} :\phi(U) \rightarrow \psi(V) $.

Any advice and maybe some examples to show how to correctly apply the definition of smooth map .

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  • $\begingroup$ I think you should study first rigorously the definition of a smooth manifold especially with examples. $\endgroup$ – TheGeekGreek Jun 22 '17 at 22:21
  • $\begingroup$ I agree with @TheGeekGreek, but in this special case you should take $\phi, \psi$ to be identity function on $U = \Bbb R^2, V = \Bbb R$. You obtain a constant function $\Bbb R^2 \to \Bbb R$ which is indeed smooth in the usual sense of term. For subset of $\Bbb R^n$, smooth manifold map coincide with the usual notion of smoothness. $\endgroup$ – user171326 Jun 22 '17 at 22:23
  • $\begingroup$ A good exercise is to show that if $(U, \phi)$ is a chart on $M$ then $\phi : U \to \phi(U)$ is smooth. $\endgroup$ – Trevor Gunn Jun 22 '17 at 23:02
  • $\begingroup$ Another good exercise is to show that if $U$ is an open subset of $\mathbf{R}^n$ then a function $f : U \to \mathbf{R}^m$ is smooth in the manifold sense if and only if it is smooth in the calculus sense. $\endgroup$ – Trevor Gunn Jun 22 '17 at 23:05
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For example, let $(a_{i,j})_{i,j = 0}^n \in \operatorname{GL}_{n+1}(\mathbf{R})$ be an $(n + 1)\times(n + 1)$ matrix. Define a map $$ f : \mathbf{RP}^{n} \to\mathbf{RP}^{n} $$ by $$ f([x^0:x^1:\dots:x^n]) = \left[ a_{0,k}x^k : a_{1,k}x^k : \cdots : a_{n,k}x^k \right] $$ where $a_{i,k}x^k$ is short for $\sum_{k = 0}^n a_{i,k}x^k$.

Recall that an atlas for $\mathbf{RP}^n$ is given by the collection of maps $$ \varphi_i : U_i := \{[x^0:\cdots:x^n] \mid x^i \ne 0 \} \to \mathbf{R}^n $$ defined by $$ \varphi_i([x^0:\cdots:x^n]) = \left( \frac{x_0}{x_i}, \dots, \widehat{\frac{x_i}{x_i}} , \dots, \frac{x_n}{x_i} \right) $$ where we omit the $i$-th entry.


Look at the maps $\phi_j \circ f \circ \phi_i^{-1}$. Let $p = [x^0 : \dots : x^n] \in U_i$ and suppose wlog that $x^i = 1$. Then \begin{align} \varphi_j \circ f \circ \varphi_i^{-1}(x^0,\dots,x^{i-1},x^{i+1},\dots,x_n) &= \varphi_j \circ f(p) \\ &= \varphi_j (\left[ a_{0,k}x^k : a_{1,k}x^k : \cdots : a_{n,k}x^k \right]) \\ &= \left( \frac{a_{0,k}x^k}{a_{j,k}x^k} , \dots, \widehat{\frac{a_{j,k}x^k}{a_{j,k}x^k}}, \dots, \frac{a_{n,k}x^k}{a_{j,k}x^k} \right) \end{align} which is defined whenever $a_{j,k}x^k \ne 0$ (which is an open set since the left hand side is continuous).

So notice, that in all of these charts, $f$ is defined by rational functions and is therefore smooth.

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  • $\begingroup$ You may wish to look at the specific example $\mathbf{RP}^1$ and take a specific $2 \times 2$ invertible matrix. $\endgroup$ – Trevor Gunn Jun 22 '17 at 22:54

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