# The composition of a smooth map with a linear isomorphism is smooth

I'm reading Jeffrey Lee's Manifolds and Differential Geometry section on manifolds with boundary.

In page 50, given an $n$-dimensional manifold $M$, he builds a smooth atlas for $\partial M$ in the following way:

Take $p \in \partial M$, then there is a chart $(U,\mathcal{x})$ such that $\mathcal{x}(p) \in \partial \mathbb{R}^n_{\lambda \geq c}$, so taking the restriction $\mathcal{x}\mid_{U \cap \partial M}$ and composing it with any linear isomorphism $F: \partial \mathbb{R}^n_{\lambda \geq c} \to \mathbb{R}^{n-1}$ we obtain a chart $(U\cap \partial M, F \circ \mathcal{x})$ for $\partial M$. The author says that with this construction we obtain smooth overlap maps but I don't see why:

Given 2 charts $(U, \mathcal{x}), (V, \mathcal{y})$ such that $p \in U\cap V$, we get the overlap map $$(F \circ \mathcal{x})\circ(F \circ \mathcal{y})^{-1} = F \circ (\mathcal{x} \circ \mathcal{y}^{-1})\circ F^{-1}$$

I know this is always a linear isomorphism, but why given any $F$ this map is always smooth?

• The composition of smooth maps is smooth. – user98602 Nov 10 '15 at 16:50
• Every linear isomorphism is smooth? – Luis Vera Nov 10 '15 at 16:52
• Differentiation is approximation of a function $f$ by a linear map. If $f$ is linear, this approximation coincides with $f$ -- there is, obviosly, no better choice. So any linear map between finite dimensional vector spaces is smooth. – Thomas Nov 10 '15 at 17:12

Let $T:\mathcal{X}\to\mathcal{Y}$ be an isomorphism vectors spaces. In order to verify that this is smooth it will be helpful to write $Tx = y$ as $$T(x_1, x_2, \dots, x_n) = (y_1(x_1, \dots, x_n), \dots, y_n(x_1, \dots, x_n))$$
with each $y_i(x_1, \dots, x_n) = \sum_{j=1}^n t_{ij}x_j$
To see that $T$ is smooth it suffices to show that each coordinate map is smooth. That is $\frac{\partial y_i}{\partial{x_j}}$ is infinitely differentiable. Observe that $\frac{\partial y_i}{\partial{x_j}} = t_{ij}$ which is constant and hence infinitely differentiable.
Overall, we could have stopped at $y_i(x_1, \dots, x_n) = \sum_{j=1}^n t_{ij}x_j$ and said $T$ is smooth because each component of its image is linear (which is smooth).