Prove $\{(x,y) \in \mathbb R^2|x^2 + y^2 > 1 \}$ is not simply connected I have literally no idea how to do this. My assignment question asks me to prove that $\{(x,y) \in \mathbb R^2|x^2 + y^2 > 1 \}$ and $\{(x,y) \in \mathbb R^2|x^2 + y^2 < 1 \}$ are homeomorphic to each other, which they can't be because only one is simply connected. So I need either a counterexample or a counter-proof to this, but I don't know how to get one as I'm not supposed to need it. Help!
I understand that one being simply connected and one not proves what I want, I just don't know how to show that that is the case.
To clarify: This is a result of a mistake in the question. We were asked to prove that there is a homeomorphism between $\{(x,y) \in \mathbb R^2|x^2 + y^2 > 1 \}$. It was only when I was researching ways to do this that I was told that because one is simply connected and one isn't, there can't be a homeomorphism. I don't want to lose out on credit for the question, but we haven't ever covered paths, simply connected, etc. in class.
 A: You've got the right idea. All you need to prove is that if $X$ is a simply connected space and $f:X\to Y$ is a homeomorphism, then $Y$ is also simply connected. If you already have that result, then there's nothing even to prove. Simply point out that only one of the two is simply connected, and you're done.
As a side note, if we add a "point at infinity" to the former set, then the two sets are homeomorphic. 
A: Do you have the following machinery in your class yet?


*

*If a space $X$ deformation retracts onto a subspace $Z$, then they have the same fundamental group.

*The fundamental group of the circle is non-trivial.
If no, I think you may need to provide more background on which tools you are supposed to use since it's not so simple to prove (2) and (2) is equivalent to your question.
If yes, try to write down explicit deformation retractions of these spaces, respectively, to a point, and to a circle (you should be able to picture them in your head; then, write down formulas using $x$, $y$, and a time variable $t$.)
A: The set $\{(x, y) \in \mathbb{R}^{2} \mid x^2 + y^2 < 1 \}$ is homeomorphic with $\mathbb{R}^{2}$ and the set $\{(x, y) \in \mathbb R^2 \mid x^{2} + y^{2} > 1 \}$ is homeomorphic with $\mathbb{R}^{2} - \mathbf{0}$. Clearly, $\mathbb{R}^{2}$ and $\mathbb{R}^{2} - \mathbf{0}$ are not homeomorphic, because to get $\mathbb{R}^{2} - \mathbf{0}$ we need to cut out a point from $\mathbb{R}^{2}$, contrary to the definition of continuous deformation. This implies that the sets in question cannot be homeomorphic.
Up to this point, I did not assume any knowledge of algebraic topology. To prove that $\{(x, y) \in \mathbb R^2 \mid x^{2} + y^{2} > 1 \}$ is not simply connected, however, we do need some algebraic topology background, because the very term simply connectedness is borrowed from algebraic topology, and not from general topology.
