Prove letter $B$ and $C$ are not homeomorphic, and $D$ and $O$ are homeomorphic

Question

This is an interesting topology question, we want to find a pair of capital letters (A - Z) that are homeomorphic and a pair that is not homeomorphic.

Homeomorphic:

The one I choose: $L$ and $T$ are homeomorphic, because I can find a function that map $L$ to $T$ (I draw it where $L = \{0\} \times [0,1] \cup [0,1] \times \{0\}$ and $T = \{0.5\} \times [0,1] \cup [0,1] \times \{1\}$, so that it's really easy to map).

But I think $D$ and $O$ are also homeomorphic since they are both closed and connected and clearly we can bend one to another. But how to map the points is my question. Could you please help?

Not homeomorphic:

My intuition makes me pick $B$ and $C$, I think if we take any point off $C$ will disconnect it, yet taking off a point from $B$ won't. Am I right? If not, could you explain what pairs you come up and why they are not homeomorphic?

Thank you in advance for your help!

Answer 1

OP requested:

Could you give a formal prove why $D$ and $O$ are homeomorphic? Can we find a function $f\colon D\to O$?

Sure.

Let $O=\{ (x,y) \mid x^2+y^2=1\}\in\mathbb{R}^2$ be the unit circle, and let $D=\{ (x,y) \mid x^2+y^2=1,\ x\geq0\}\cup\{(0,y)\mid -1\leq y\leq 1\}\in\mathbb{R}^2$ be the union of the half circle and the line segment.

Define a map $f\colon D\to O$ by $$ f\bigl((x,y)\bigr) = \begin{cases} (x,y) & \text{if $x>0$} \\ (-\sqrt{1-y^2},y) & \text{if $x=0$} \end{cases} $$ and its inverse $g\colon O\to D$ by $$ g\bigl((x,y)\bigr) = \begin{cases} (x,y) & \text{if $x\geq0$} \\ (0,y) & \text{if $x\leq0$} \end{cases} $$

Then it is easy to check that $f$ and $g$ are continuous, $f\circ g=\mathrm{id}_{O}$, and $g\circ f=\mathrm{id}_{D}$.

Answer 2

B and C are not homeomorphic because B has no cut points
and every point of C except the end points are cut points

L and T are not homeomorphic because T has a point
whose removal results in 3 components and L does not.

Yes, D and O are homeomorphic.