Smooth map from a manifold to the real numbers

Question

In setting up the definition for the tangent vector space at a point in a differentiable manifold, Carroll references the space of smooth functions $f: M \to \mathbb{R}$. What does it mean for a function from the manifold to the real numbers to be smooth? I apologize if this is straightforward or in the text, but I didn't see it and it is pretty fundamental so I want to make sure I understand it correctly. This is on Carroll Spacetime and Geometry page 63

Answer 1

Take a (real) differentiable n-dimensional manifold $M$.

By definition, you can choose subsets $M_i$ of $M$ such that $M=\cup_i M_i$ and injective functions $\phi_i:M_i\rightarrow V_i$ with $V_i$ open conected subsets of $\mathbb{R}^n$. Keep in mind that because the $\phi_i$'s are injective, you can find an "inverse" $\phi_i$ (which domain is the image of the fucntion $\phi_i$) such that the composition $\phi_i^{-1}\circ \phi_i$ is the identity function of $M$. WLOG suppose that these functions are bijections and $V_i=\mathbb{R}^n$.

With arrows:

\begin{equation} \mathbb{R}^n \rightarrow M_i\subset M \rightarrow \mathbb{R} \end{equation}

The left arrow is $\phi_i^{-1}$ and the right arrow is $f$.

Then $f:M\rightarrow \mathbb{R}$ is smooth if the composition $f\circ\phi_i^{-1}:\mathbb{R}^n\rightarrow\mathbb{R}$ is smooth in the usual sense.