# Show that an open subset of a smooth manifold is again smooth.

This step is essential in proving that $GL(n,\mathbb{R})$ is a smooth manifold. I already proved that $GL(n,\mathbb{R})$ is an open subset of $M_n(\mathbb{R})$. Can you help me prove the above statement? Thanks!

p.s. My idea (but I'm not sure) is that we will consider the smooth structure of the smooth manifold and restrict the corresponding maps to the intersection of the open set and the corresponding open sets from the atlas.

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$M_n$ is a vectorspace, hence isomorphic and hence diffeomorphic to $\mathbb{R}^k$. An open subset $U$ is quite obviously a manifold -- take $(U,id)$ as an atlas. –  user20266 Aug 13 '12 at 8:13
Your idea is correct. –  Alexander Thumm Aug 13 '12 at 8:20
@Alexander, so is it considered a correct proof? :) –  John Thompson Aug 13 '12 at 8:35
@JohnThompson: Well, you would still have to write it out :D –  Alexander Thumm Aug 13 '12 at 8:42
Any open subset $V$ of a smooth manifold $M$ is also a smooth manifold . If $\{(U_{\alpha},\phi_{\alpha})\}$ is an atlas for $M$, Then $\{(U_{\alpha}\cap V,\phi_{\alpha}|_{U_{\alpha}\cap V}\}$ is an atlas for $V$ , where $\phi_{\alpha}|_{U_{\alpha}\cap V}:U_{\alpha}\cap V\rightarrow \mathbb{R}^n$ denotes the restriction of $\phi_{\alpha}$ to the subset $U_{\alpha}\cap V$.
for any positive integer $m,n$ let $\mathbb{R}^{m\times n}$ be the vector space of all $m\times n$ matrices . Since $\mathbb{R}^{mn}\cong \mathbb{R}^{m\times n}$ , we give it the topology of $\mathbb{R}^{mn}$. As you have already shown that $GL_n(\mathbb{R})$ is an open subset of $\mathbb{R}^{n^2}\cong M_n(\mathbb{R})$ so is a again a manifold.