I need to show that given $x,y\in M, x\neq y$ where $M$ is a smooth manifold, $\exists$ a smooth function $f:M\longrightarrow \mathbb{R}$ such that $f(x)=0, f(y)=1.$

Attempt of the proof: Since $M$ is a smooth manifold, there exists an atlas $\{(U_\alpha,\phi_ \alpha)\}_{\alpha\in\mathbb{N}}$ such that all transition maps are smooth. Take two charts $(U_1,\phi_1),(U_2,\phi_2)$ such that $x\in U_1,y\in U_2$ and $U_1\cap U_2\neq \emptyset$, where $\phi_1:U_1\longrightarrow V_1,\phi_2:U_2\longrightarrow V_2$ are homeomorphisms, and $V_1,V_2\subset \mathbb{R^n}$ are open. By the definition of smooth manifold, I know that the transition map $$\phi_2 \phi_1^{-1}:\phi_1(U_1\cap U_2)\subset \mathbb{R^n}\longrightarrow \phi_2(U_1\cap U_2)\subset \mathbb{R^n}$$ is smooth. I now need to create a smooth map $f:M\longrightarrow \mathbb{R}$ that satisfies the required property.

In my definition $f:M\longrightarrow \mathbb{R}$ is smooth if for any coordinate chart $(U,\phi)$, $f\phi^{-1}:\phi(U)\subset \mathbb{R^n}\longrightarrow\mathbb{R}$ is smooth.

I cannot proceed ahead.

EDIT: One natural thing that I could do in this case is to send $U_2$ to $V\subset\mathbb{R}$ by a homeomorphism. I think that is not possible, especially if the dimension of the maniold is $> 1$. If that was the case, the transition map is smooth, but still would not fulfill the requirement of my smooth function $f.$


Let $(U,\phi)$ be a chart around $y\in U$ such that $x\not\in U$ and let $V=\phi(U)\subset \mathbb{R}^n$. Now choose a smooth and compactly supported function $g\in C_c^\infty(V)$ such that $g(\phi(y))=1$. This is possible, just take a variant of a standard mollifier. This gives you a smooth, compactly supported function (smooth as composition of smooth maps)

$$f=g\circ\phi:U\rightarrow \mathbb{R}$$

Because the function has compact support, you can just extend it by $0$ to all of $M$ and it will still be smooth. This gives you a smooth function $f:M\rightarrow\mathbb{R}$ such that $f(x)=0$ and $f(y)=1$.


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