Showing $S^1$ is not homeomorphic to $\mathbb R^2$ with the subspace topology So I want to say we let $f:S^1→\mathbb R^2$ be a homeomorphism, and since we can find a bijection, I want to say we can open an open set in the plane using the standard product topology, and take its pre-image, which is assumed to be open by continuity of f but this is not always the case in $S^1$. 
Help or hint please. 
 A: At least two proofs come to mind. 


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*$S^1$ is compact, and $\mathbb R^2$ is not, but any continuous map takes compact sets to compact sets, so $\mathbb R^2$ cannot be the image under any continuous map of $S^1$ with the given topologies.

*$S^1$ minus any two points is disconnected, but $\mathbb R^2$ minus any two points remains connected. This forms a proof as follows : consider $p,q \in \mathbb R^2$  ,then the set $\mathbb R^2 \setminus \{p,q\}$ is a connected (open) set, hence its image under any injective continuous map $f$ to $\mathbb S^1$ should be connected, but the image is $\mathbb S^1 \setminus \{f(p),f(q)\}$ which is disconnected, since $f(p) \neq f(q)$ by injectivity of $f$. Therefore, no injective continuous map from $\mathbb R^2$ to $S^1$ exists in the given topologies.
Also see if something like this works out (exercise) : $S^1$ is the quotient space of $[0,1]$ under the identification $a \sim b \iff a=b $ or $\{a,b\} = \{0,1\}$. Therefore, any continuous map $f$ from $S^1$ to $\mathbb R^2$ lifts to a continuous map $\psi$ from $[0,1]$ to $\mathbb R^2$, which satisfies $\psi(0) = \psi(1)$, via $\psi(a) = f([a])$.  The reverse is also true : any such continuous map $\phi : [0,1] \to \mathbb R^2$ with $\phi(0) = \phi(1)$ factors to a continuous map given by $g([a]) = \phi(a)$. So, we have an identification between continuous maps from $S^1$ to $\mathbb R^2$, and from $[0,1]$ to $\mathbb R^2$.
Using the fact that $[0,1]$ is a bounded interval, see if you can make any argument using the above identification  to show that $S^1  \not \cong \mathbb R^2$. This is a more comfortable setting to try out your "direct method" that you were attempting in the post.
