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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$\begingroup$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.
$\endgroup$ $\begingroup$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 $\mathbb{R}^2$.
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