Continuous Injective Map $f :\mathbb{S}^2 \rightarrow \mathbb{S}^1$

Question

I'm trying to decide whether if there exists a continuous injective map $f :\mathbb{S}^2 \rightarrow \mathbb{S}^1$ (in the Euclidean topology) by elementary topology arguments. I first tried to prove that there exist no bijective such map by connectedness arguments. It is a well known fact that $\mathbb{S}^2$ is a connected subset of $\mathbb{R}^3$. In fact, let $P \in \mathbb{S}^2$ be an arbitrary point of the sphere, it follows that $\mathbb{S}^2 \setminus \{P\}$ is also a connected set. Suppose it exists a bijective continuous map $f:\mathbb{S}^2 \rightarrow \mathbb{S}^1$, then $f(\mathbb{S}^2 \setminus \{P\})$ must be a connected subset of $\mathbb{S}^1$. As $f$ is bijective, let $Q=f(P)$, then $f(\mathbb{S}^2 \setminus \{P\}) = \mathbb{S}^1 \setminus Q$. However, $\mathbb{S}^1 \setminus Q$ cannot be connected; hence, such function cannot exist.

When relaxing the bijective hypothesis the same argument cannot be used and I don't know how to proceed; it is trivial to find a function $f:\mathbb{S}^1 \rightarrow \mathbb{S}^2$ under this conditions but I can't think of any interchanging the domain and range. I would thank any hint or heuristics to solve the problem.

Answer 1

As Lee Mosher comments, there are some similar questions in math.stackexchange. Using the methods described in their answers, let us assume that $f : S^2 \to S^1$ is injective and proceed as follows:

Case 1. $f(S^2) \ne S^1$. Pick $x \in S^1 \setminus f(S^2)$. There exists a homeomorphism $h : S^1 \setminus \{x\} \to \mathbb R$ and we get an continuous injection $g = h \circ f : S^2 \to \mathbb R$. The image $J = g(S^2)$ is a compact connected subset of $\mathbb R$, thus a closed interval $[a,b]$ which must be homeomorphic to $S^2$ via $G : S^2 \stackrel{g}{\to} [a,b]$. Let $c \in (a,b)$. But $[a,b] \setminus \{c\}$ is not connected, thus also $S^2 \setminus \{G(c)\}$ is not connected. This is false, thus case 1 is impossible.

Case 2. $f(S^2) = S^1$. Then $f$ is a continuous bijection between compact Hausdorff spaces, hence a homeomorphism. It restricts to a homeomorphism $F : S^2 \setminus \{northpole\} \to S^1 \setminus \{f(northpole)\}$. Its domain is homeomorphic to $\mathbb R^2$, its range homeomorphic to $\mathbb R$. Thus we get a homeomorphism $h : \mathbb R^2 \to \mathbb R$. A similar argument as in case 1 (remove a point of $\mathbb R$) leads to a contradiction. Thus also case 2 is impossible.