# $U\subset \mathbb{R}^d$ open connected, $\varphi:U\to \mathbb{R}^m$ differentiable, injective, regular. Is $\varphi\left (U\right )$ a manifold?

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?

• Try to construct a counterexample like this: upload.wikimedia.org/wikipedia/commons/9/90/… – Neal Nov 10 '18 at 18:45
• 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. – Laz Nov 10 '18 at 18:56
• Thank you, @Neal . This one seems to work: math.stackexchange.com/questions/2067540/… – solomeo paredes Nov 10 '18 at 19:09