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I am currently trying to prove that for differentiable manifolds $M \subset \mathbb{R}^m$ and $N \subset \mathbb{R}^n$. It holds that: $T_{(x,y)} (M \times N) = T_x M \times T_y N$ for arbitrary $x \in M$ and $y \in N$. I have seen proves see here or here. However, I do not see why this is necessary. Our definition of the tangent space is: Let $M\subset \mathbb{R}^n$ be a $k$-dimensional manifold and $ x\in M$. Then $T_x = \{c'(0) : c(0)=x, c: I \to M\}$ where $I$ is an open interval containing $0$ and $c$ is continuously differentiable.

Using this definition we would obtain: $ T_{(x,y)}(M \times N) = \{c'(0) : c(0)=(x,y), c: I \to M\times N \} = \{(c_M'(0),c_N'(0)) : c_M(0)=x, c_N(0) = y, c_M: I \to M, c_N: I \to N\} = \{c_M'(0) : c_M(0)=x, c: I \to M\} \times \{c_N'(0) : c_N(0)=y, c: I \to N\} = T_xM \times T_yN$

Is there something I am missing or why do the other proofs use linear isomorphisms.

user007
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1 Answers1

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Your idea is basically equivalent to the ones in those proofs. They’re just being a bit more explicit/careful/formal with the projection maps and embeddings.

In your case, you’re implicitly using maps $\pi_M:M\times N\to M$ and $\pi_N:M\times N\to N$ to get $c_M=\pi_M\circ c$ and $c_N=\pi_N\circ c$.

Note that we care about linear maps here because we want to show that the two spaces are equivalent as vector bundles over $M\times N$

user815293
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