Immersion of sphere into $\mathbb{R}^2$

Question

This is the first time I ask a question, and as a French I will probably make a few mistakes. Sorry for that.

I want to show that there are no immersions between sphere and plane (in particular, I want to show that there is no immersion between $S^2 = \{ x^2+y^2+z^2=1 \}$ and $\mathbb{R}^2$).

I read somewhere online this very concise proof (from what looked like a trustable source, in a good university pdf course), but I don't understand it.

"Let $f$ be an immersion from $S^2$ to $\mathbb{R}^2$. The image of $f$ is open (since an immersion between two spaces of same dimension is an open map) and closed (by compactness). Thus this is $\mathbb{R}^2$ (by connectedness), which is absurd, by compactness."

Can you help me figure it out? Especially the first part : how can the argument of $f$ being an open map can be used to show that the image is open since $S^2$ is closed?

Answer 1

$S^2$ is indeed a closed subset of $\mathbb R^3$. However, when you consider an immersion $f: S^2 \mapsto \mathbb R^2$, you're dealing with $S^2$ as a topological space on its own. Therefore as for all topological space $X$, $X$ itself is open. That is an axiom of a topological space.

Answer 2

Since $S^2$ is compact, the image $f(S^2)$ is a compact subset of $\mathbb R^2$ (this is true for any continuous $f : S^2 \to \mathbb R^2$). Hence $f(S^2)$ is a closed subset of $\mathbb R^2$.

Now $f$ is an immersion. This means that the rank of the differential $df_x : T_xS^2 \to T_{f(x)} \mathbb R^2$ is $2 = \dim S^2$ for all $x \in S^2$. Since also $\dim \mathbb R^2 = 2$, this implies that there exists an open neigborhood $U$ of $x$ in $S^2$ such that $f$ maps $U$ diffeomorphically onto an open $V \subset \mathbb R^2$. We conclude that $f(S^2)$ is open in $\mathbb R^2$.

Therefore $f(S^2)$ is an open and closed nonempty subset of $\mathbb R^2$. Since $\mathbb R^2$ is connected, we must have $f(S^2) = \mathbb R^2$. This is a contradiction because $f(S^2)$ is compact.