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I am dealing with the following exercise:

Ley $d,m\in \mathbb{N}$, $d\leq m$ and $U\subset \mathbb{R}^d$ an open connected subset. Let $\varphi:U\to \mathbb{R}^m$ be a differentiable, injective and regular (i.e. the Jacobian matrix $d\varphi_p\in \mathbb{R}^{m\times d}$ has rank $d$ for every $p\in U$). Is it true that $X=\varphi \left (U\right )$ is a manifold in $\mathbb{R}^m$?

This was my attempt: if we intend to prove the positive assertion, $X$ may be a manifold which can be covered with just one chart, where our candidate for chart is $\varphi$. We only have to prove that $\varphi$ is an homeomorphism.

Fix $p\in U$ and consider $v_1,\cdots ,v_{m-d}\in \mathbb{R}^m$ such that they extend the $d$ columns of $d\varphi_p$ to a basis for the vector space $\mathbb{R}^m$. Define $F:U\times \mathbb{R}^{m-d}\to \mathbb{R}^m$ as $F\left (x,t\right ):=\varphi \left (x\right )+\sum_{i=1}^{m-d}t_iv_i$. Then $dF_{\left (p,0\right )}=\left (d\varphi_p\mid v_1\mid \cdots \mid v_{m-d}\right )$ is a non-singular matrix in $\mathbb{R}^{m\times m}$. We apply the inverse function theorem and get an open subset $A\subset U\times \mathbb{R}^{m-d}$ containing $\left (p,0\right )$ and $B\subset \mathbb{R}^m$ such that $F:A\to B$ is a diffeomorphism. Now take the projection $\pi:U\times \mathbb{R}^{m-d}\to U$. Since $\left (p,0\right )\in A$ and $A$ is open, then $\left (x,0\right )\in A$ for every $x$ sufficiently close to $p$. Therefore, for those values for $x$, we have $F\left (x,0\right )=\varphi \left (x\right )$, so $x=\pi F^{-1}\varphi \left (x\right )$. We found a local left-inverse for $\varphi$ near an arbitrarily chosen $p\in U$. We need an actual inverse, and we need to define it globally. How would you do it, providing that it actually exists? Notice that I have not made use of the connectedness of $U$ yet.

If not, then I suppose that the assertion of the exercise is false. Which counterexample would you provide?

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    $\begingroup$ Try to construct a counterexample like this: upload.wikimedia.org/wikipedia/commons/9/90/… $\endgroup$ – Neal Nov 10 '18 at 18:45
  • $\begingroup$ Yeah, I didn't realize it's missing the hypothesis that $\varphi$ is homeomorphism onto its image. So, it's false and the counterexample @Neal provided is correct. By the way, nice name Solomeo Paredes, lol. $\endgroup$ – Laz Nov 10 '18 at 18:56
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    $\begingroup$ Thank you, @Neal . This one seems to work: math.stackexchange.com/questions/2067540/… $\endgroup$ – solomeo paredes Nov 10 '18 at 19:09

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