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

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.

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.