What's going on with the quotient group $SO(2,\mathbb{R})/\langle-I\rangle$ and the endomorphism of $SO(2,\mathbb{R})$ given by $A\mapsto A^{2}$? I've been learning abstract algebra and I'm at orthogonal groups, and wishing to solidify my shaky intuition about quotients I wanted to look at quotients of the special orthogonal group $SO(2,\mathbb{R})$ given its comfortable structure. I considered the quotient $SO(2,\mathbb{R})/\langle-I\rangle$, where $I$ is the identity matrix, and I was immediately perplexed.
So, the elements of this group can be thought of as antipodal points on the unit circle, the operation given by the usual rotation by $\theta$ except rotating by $\theta=\pi$ is an identity, or rather, thinking of $SO(2,\mathbb{R})$ as being the quotient $\mathbb{R}/2\pi\mathbb{Z}$, $SO(2,\mathbb{R})/\langle-I\rangle$ is the quotient $\mathbb{R}/\pi\mathbb{Z}$, right? However, by that note, shouldn't $SO(2,\mathbb{R})/\langle-I\rangle$ simply be isomorphic to $SO(2,\mathbb{R})$? Thinking about it geometrically, both groups "wrap around" in an identical manner, that is they both consist of rotations, where it follows that if you rotate by a large enough angle the overall effect is as if you did nothing.
However, if you think of $SO(2,\mathbb{R})/\langle-I\rangle$ as the image of the endomorphism $f:SO(2,\mathbb{R})\to SO(2,\mathbb{R})$ given by $A\mapsto A^{2}$ with the first isomorphism theorem, there's no way $SO(2,\mathbb{R})/\langle-I\rangle$ is isomorphic to $SO(2,\mathbb{R})$ since the kernel of $f$ is $\langle-I\rangle$! I find this most bewildering, that these two groups that seem basically identical can't be isomorphic. Where is my thinking incorrect here, and how should I be comparing these two groups? I know that $SO(2,\mathbb{R})$ is a so-called "lie group", I have not formally studied the subject of lie groups but I will nonetheless add to the tags accordingly.
 A: 
However, if you think of $SO(2,\mathbb{R})/\langle-I\rangle$ as the kernel of the endomorphism $f:SO(2,\mathbb{R})\to SO(2,\mathbb{R})$ given by $A\mapsto A^{2}$ with the first isomorphism theorem

It's not the kernel, it's the cokernel. The kernel is $\{ \pm I \}$, which you correctly state later, so I'm not sure what you mean here. And yes, $SO(2) / \{ \pm I \}$ is just $SO(2)$ again. More generally, for every $n$, the map $z \mapsto z^n$ defines a short exact sequence
$$1 \to \mathbb{Z}/n\mathbb{Z} \to S^1 \xrightarrow{z \mapsto z^n} S^1 \to 1$$
where $S^1 \cong SO(2)$ and $\mathbb{Z}/n\mathbb{Z}$ is embedded in $S^1$ as the group of $n^{th}$ roots of unity (thinking of $S^1 = \{ z \in \mathbb{C} : |z| = 1 \}$). This is a version of the Kummer exact sequence.
A: There is no contradiction in these groups being homeomorphic and isomorphic. The original $SO(2)$ is isomorphic to $([0,2\pi]/\sim,+)$ (the boundaries $0$ and $2\pi$ identified). And the other is isomorphic to $([0,\pi]/\sim,+)$.  They are, in essence, the same, we are only "changing the scale".
