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Let us call surface patch a smooth function $x:U\to\mathbb R^3$ with domain an open set in $\mathbb R^2$ that is injective and has linearly independent partial derivatives at each point of its domain.

The image of such a thing can be closed. For example, the function $x:(-\pi/2,3\pi/2)\times\mathbb R\mapsto\mathbb R^3$ with $$x(\theta,t)=(\cos\theta,\sin2\theta,t)$$ for all $(\theta,t)$ in the domain of $x$ works. Here is a picture of the image of $(-\pi/2,3\pi/2)\times[0,1]$:

enter image description here

This example is constructed from the curve $$\gamma:\theta\in(-\pi/2,3\pi/2)\mapsto(\cos t,\sin2t)\in\mathbb R^2$$ by adding one extra coordinate that does not see any action. This curve has the property that its image is closed.

enter image description here

In principle, we can wrap the $t$ direction in the surface patch $x$ so that we also have such a flowery shape, and obtain a surface patch with compact image… but it seems complicated to do.

Does anyone have an example of a surface patch with compact image that is simpler than this?

Mariano Suárez-Álvarez
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    I'm a little bit confused on your definition. You say a surface patch must be an injective mapping and smooth mapping from an open set. I don't see how a compact image from an open domain is possible under those conditions. Even if you allow noninjectivity at boundaries, the example you provide is not injective for many different lines. – Ninad Munshi May 15 '23 at 07:42
  • There are no boundaries in the domain of a surface patch, since it is an open set. In what lines is my example non-injective? – Mariano Suárez-Álvarez May 15 '23 at 08:02
  • If you want to see how the image can be compact, simply consider the curve I described in the question: it is a function $(0,2\pi)\to\mathbb R^2$ which is injective, has non-zero derivative everywhere, and its image is a compact set. – Mariano Suárez-Álvarez May 15 '23 at 08:03
  • Your curve $\gamma \colon (0, 2\pi) \to \mathbb{R}^3, t \mapsto (\cos(t), sin(2t))$ is not injective because $\gamma(\pi/2) = \gamma(3\pi/2) = (0,0)$. Its image is not open because it has limit point $(1,0)$ which is not in the original set (because $0$ and $2\pi$ are excluded from the domain). There can be no such continuous curve from an open set because preimages of closed sets under continuous functions are closed. – SV-97 May 15 '23 at 08:40
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    I think the author meant $(\sin(t),\sin(2t))$, i.e. a leminiscate (which is injective and has compact image) – porridgemathematics May 15 '23 at 08:41
  • The interval in the domain of the curve/patch was wrong. That's indeed the curve, @porridgemathematics, but I was just using a different parametrization. – Mariano Suárez-Álvarez May 15 '23 at 08:45
  • @SV-97, indeed, the image is not open, but it is not needed to be open, and I don't really know what you mean when you say that there can be no such continuous curve. – Mariano Suárez-Álvarez May 15 '23 at 08:47
  • @MarianoSuárez-Álvarez the first thing that comes to my mind (for a flower shape): consider the torus $S^1 \times S^1$ embedded in a standard way into $\mathbb{R}^3$, then take a torus knot $S^1 \rightarrow S^1 \times S^1$, say $g: z \rightarrow (z^m,z^n)$ for positive integers $m,n$, with $\gcd(m,n) \neq 1$, so that on an open interval of $(0,2 \pi)$, there is an injection $(0,2 \pi) \rightarrow_{\exp(i \theta)} \rightarrow_{g} S^1 \times S^1$, then let $\hat{n}$ be the unit outer normal to the curve and use this to parameterize the closed flower in the perpendicular direction. – porridgemathematics May 15 '23 at 09:12
  • @SV-97 I think you mean to say $Y \subset \mathbb{R}^n$ is closed, and $X \subset \mathbb{R}^m$ , $m < n$ is open.. otherwise you are trying to prove there can be no continuous function $f : X \rightarrow Y$ such that $X$ is open, and $Y$ is closed, which doesn't make sense because topological spaces are always open/closed with respect to their own topology. Also , this is false, take $(-\pi,\pi) \rightarrow \mathbb{R}^2 ; t \rightarrow (\sin(t),\sin(2t))$, the image is compact, and this is an injective immersion. – porridgemathematics May 15 '23 at 09:40

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