$\Psi$ is a morphism of ringed spaces iff it is smooth in local coordinates

Question

Here is what I have to show:

Let $(M,\mathcal{F})$ and $(N,\mathcal{F}')$ be smooth manifolds of class $C^{\infty}$ and let $\Psi:M\to N$ be a continuous map. Show that the following conditions are equivalent:

1. In local coordinates, $\Psi$ is given by a smooth map.
2. $\Psi$ is a morphism of ringed spaces.

Definitions (manifolds are assumed Hausdorff):

• Let $(M,\mathcal{F})$ be a topological manifold $M$ (i.e. a topological space locally homeomorphic to $\mathbb{R}^{m}$) equipped with a sheaf of functions $\mathcal{F}$. It is said to be a smooth manifold of class $\mathcal{C}^{\infty}$ if every point has a neighbourhood isomorphic to the ringed space $(\mathbb{R}^{n},\mathcal{G})$ where $\mathcal{G}$ is a ring of functions of this class.
• A coordinate system on an open subset $U$ of a smooth manifold $(M,\mathcal{F})$ is an isomorphism between $(U,\mathcal{F})$ and an open subset in $(\mathbb{R}^{m},\mathcal{G})$ where $\mathcal{G}$ are functions of the same class on $\mathbb{R}^{m}$.

1. $\implies$ 2.

My first problem is what does "In local coordinates, $\Psi$ is given by a smooth map" mean? When I first learned Differential Geometry, it used to mean that for some appropriate charts $(U,\phi)$ and $(V,\varphi)$ of $M$ and $N$, we had $$\varphi\circ\Psi\circ\phi^{-1}\vert_{\phi\left(\,\Psi^{-1}(V)\cap U\,\right)}:\mathbb{R}^{m}\to\mathbb{R}^{n}\in C^{\infty}$$ I do not understand how to make a link between the two definitions if they are indeed equivalent.

This problem left aside, if I take the "new" definition and if I understand correctly, $\Psi$ has to be smooth for some coordinate system on $V$, an open subset in $N$? Let $\Lambda_{V}$ be a coordinate system on $V$.

I need to prove that $\Psi$ is a morphism of ringed spaces, i.e. $\forall\,V\,\text{open}\subset N,\,\forall\,f\in\mathcal{F}'(V)$ we have $$f\circ\Psi\in\mathcal{F}(\Psi^{-1}(V))$$

For all $f\in\mathcal{F}'(V)$, we have $f\circ\Lambda_{V}^{-1}\in\mathcal{G}(\Lambda_{V}(G))$ since $\Lambda_{V}^{-1}$ is well-defined and is a morphism of ringed spaces: $$\Lambda_{V}^{-1}:(\Lambda_{V}(V),\mathcal{G}(V))\to(V,\mathcal{F}'(V))$$

We also have

$$\Lambda_{V}\circ\Psi:\Psi^{-1}(V)\subset M\to\Lambda_{V}(V)\subset\mathbb{R}^{n}$$

But now, if it is the right way to begin, how can I go further?

Answer 1

My first problem is what does "In local coordinates, $\displaystyle\Psi$ is given by a smooth map" mean?

By your definition, $U$ is a coordinate system of $(M,\mathcal{F})$ if $(U,\mathcal{F}_{|U})$ is isomorphic as ringed space to $(\mathbb{R}^m,\mathcal{C}^{\infty}_m)$, where $\mathcal{C}^{\infty}_m$ is the sheaf of smooth functions on $\mathbb{R}^m$; then, your hypothesis is: for all coordinate systems $U$ of $(M,\mathcal{F})$ and $V$ of $(N,\mathcal{G})$, $(\Psi_{U,V},\Psi_{U,V}^{\sharp}):\left(U,\mathcal{F}_{|U}\right)\to\left(V,\mathcal{G}_{|V}\right)$ is a smooth map; that is, let $(\varphi_U,\varphi_U^{\sharp}):\left(U,\mathcal{F}_{|U}\right)\to(\mathbb{R}^m,\mathcal{C}^{\infty}_m)$ be the isomorphism of ringed spaces between $U$ and $\mathbb{R}^m$ and let $(\varphi_V,\varphi_V^{\sharp}):\left(V,\mathcal{G}_{|V}\right)\to(\mathbb{R}^n,\mathcal{C}^{\infty}_n)$ be the isomorphism of ringed spaces between $V$ and $\mathbb{R}^n$, $\varphi_V\circ\Psi_{U,V}\circ\varphi_U^{-1}:\mathbb{R}^m\to\mathbb{R}^n$ are smooth maps!

You can check on the stalks that: $$ \forall x\in U,\,\varphi_{U,x}^{\sharp}\circ\Psi_{U,V,\Psi_{U,V}(x)}^{\sharp}\circ\varphi_{V,\varphi_V(\Psi_{U,V}(x))}^{\sharp}:\mathcal{C}^{\infty}_{n,\varphi_V(\Psi_{U,V}(x))}\to\mathcal{C}^{\infty}_{m,\varphi_U(x)} $$ are well defined morphism of (local) rings, and therefore \begin{equation} (\varphi_V\circ\Psi_{U,V}\circ\varphi_U^{-1},(\varphi_V\circ\Psi_{U,V}\circ\varphi_U^{-1})^{\sharp}):(\mathbb{R}^m,\mathcal{C}^{\infty}_m)\to(\mathbb{R}^n,\mathcal{C}^{\infty}_n) \end{equation} is a morphism of (locally) ring spaces.

From all this, you can conclude that $\Psi_{U,V}$ are morphisms of ringed spaces for any $U$ and $V$; and easily you can note that this reasoning is invertible, that is $\Psi$ is a smooth map if and only if $(\Psi,\Psi^{\sharp})$ is a morphism of (locally) ringed space.