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Sorry if this is a vague question.

I was handed a map between two manifolds, say, $M\to \mathbb{R^n}$. I was indeed able to find an open cover of $M$ and show that there is a homeomorphism between each open set and some open subset of $\mathbb{R^n}$. Proving homeomorphism usually takes me a long time but it's usually doable. But when it comes to showing that each chart is $C^{\infty}$-compatible, in other words, every transition map is smooth, I'm getting nowhere.

The same situation again happened on proving that Grassmannian in the case of $G(2,4)$ has a smooth structure. I identified each open set that covers $G(2,4)$ and proved homeomorphism, but then when it comes to proving the smoothness of transition maps, I get nowhere again. I get the map between two set of matrices and I had no idea how to argue. I think this happens because general theorems on homeomorphism are well-known but smoothness is really different for every situation. But since I am really having trouble proving smoothness, I'd like to know if there is any standard way or theorem of showing that the transition map is smooth. Thank you for your answers.

able20
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    You can identify $n\times m$ matrices with $\mathbb{R}^{nm}$ so your map is really a map $\mathbb{R}^{nm} \to \mathbb{R}^{nm}$ which is smooth if and only if its component functions are smooth. In terms of the matrices, you can write the entries of the output matrix as functions of the entries of the input matrix. The map is smooth if and only if those functions are smooth. – Michael Albanese Feb 08 '21 at 13:01

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Well, if you're working with $\mathbb{R}^n$, smoothness just means that every component is smooth in all the others (due to some MVT argument, it even suffices to be smooth in every variable separately). And the next trick is that everything you are doing in the Grassmann example is rational (as in: your transition functions are ultimately rational functions), and rational functions are always smooth (where defined).

philip98
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