If $\Omega$ is a simply-connected domain in $\mathbb R^n$ and $f$ is a injective continuous map from $\Omega$ to $\mathbb R^n$, then is it necessary that $f(\Omega)$ a simply-connected domain?

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    $\begingroup$ Brouwer's domain invariance theorem applies and concludes that $\Omega$ is homeomorphic to $f(\Omega)$. In particular, if one is simply connected the other must also be. $\endgroup$ – hmakholm left over Monica Oct 25 '11 at 13:32
  • $\begingroup$ I guess by domain it's meant an open, connected set? If so, to add a little, a homeomorphism between spaces gives you an isomorphism between the respective fundamental groups, so that the zero group is sent to the zero group by $f_*$. $\endgroup$ – gary Oct 27 '11 at 6:12

This is true by invariance of domain. Since $f$ is injective and continuous, its inverse is a continuous map from $f(\Omega)$ to $\Omega$. Thus, any path $\gamma$ in $f(\Omega)$ can be lifted to a path $\gamma'=f^{-1}\circ\gamma$ in $\Omega$. Since $\Omega$ is simply connected, there is a continuous map $g:[0,1]^2\to\Omega$ that shrinks $\gamma'$ to a point, and since $f$ is continuous, this yields a continuous map $f\circ g$ which shrinks $\gamma$ to a point.

As Henning has pointed out in a comment, you can skip the details if you know that being simply connected is a topological invariant.


It is worth noting that a very slight modification, where we allow the dimension of the ambient space to be different, makes the answer "no".

If $\Omega$ is a simply-connected domain in $\mathbb{R}^n$ and $f$ is an injective continuous map from $\Omega$ to $\mathbb{R}^m$, then $f(\Omega)$ is not necessarily a simply-connected domain.

An example is given by the "figure 6", thought of as a map from an open interval in $\mathbb{R}^1$ into $\mathbb{R}^2$.

  • $\begingroup$ I don't think this is right. Firstly, the proof in the other answer does not require the dimensions to be the same. Secondly, the "figure 6" map is not injective. $\endgroup$ – Doris Nov 27 '19 at 16:56

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