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Let $f:M\to N$ be a smooth bijection between manifolds with same dimension. Do we necessarily have

$$df_p(T_pM)=T_{f(p)}N.$$

I think it is probably not true. But I can't give a counterexample...

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  • $\begingroup$ Is this question the same as "Are smooth bijections of manifolds of the same dimension submersions?" ? $\endgroup$ – Selene Auckland Jul 24 at 7:54
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The map $f:\mathbb{R}\to\mathbb{R}$ given by $f:x\mapsto x^3$ is a smooth bijection but $df_0(T_0\mathbb{R}) = 0\neq T_{f(0)}\mathbb{R}$.

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  • $\begingroup$ Is this question the same as "Are smooth bijections of manifolds of the same dimension submersions?" please? $\endgroup$ – Selene Auckland Jul 24 at 7:52
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    $\begingroup$ @SeleneAuckland Let me answer your question with a question: is the function $f$ in my answer a smooth bijection between two manifolds of the same dimension that is not a submersion? :) $\endgroup$ – Neal Jul 24 at 12:59
  • $\begingroup$ Thanks. I guess, yes that $f$ is such. I was just wondering why OP didn't use the word "submersion". So it is the same question then? $\endgroup$ – Selene Auckland Jul 25 at 2:00
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How about $f: \mathbb{R} \to \mathbb{R}$ by $f(x)=x^3$? Certainly smooth, certainly bijective. However, $df_0$ will map the $1$-dimensional tangent space to a point. The root issue is that a smooth bijection can still have critical points.

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