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I've got here an exercise that says: "Show that the map $F:\mathbb{R}\rightarrow \mathbb{R^2}$ defined by $F(t)=(\cos t, \sin t)$ is an immersion". $F$ is an immersion if $dF_x:T_x\mathbb{R}\rightarrow T_{F(x)}\mathbb{R^2}$ is injective. Now $dF_x$ is $(-\sin x, \cos x)^t$ (am I wrong?) and it isn't injective, so it isn't an immersion. Is it correct?

Martin Brandenburg
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batman
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  • Fixing each $x$, $(-\sin x, \cos x)^t$ as a linear map is injective. –  Nov 04 '13 at 18:36

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Note that for all $x \in \mathbb R$ $$\text{rank } (-\sin x,\cos x)^T=1 $$since $\sin x$ and $\cos x$ don't have common zeros.

user1337
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    Ok, I think I was a bit confused. So, every time I have to check if a map $F$ is an immersion I compute its Jacobi matrix, because $dF_p$ is represented by the Jacobi matrix, and check that its rank is equal to the number of columns. The same thing if I want to check if $F$ is a submersion, but this time I have to check that the rank of the Jacobi matrix is equal to the number of rows. Right? – batman Nov 05 '13 at 15:27
  • @Claretta correct! – user1337 Nov 05 '13 at 16:21