# Which letter is not homeomorphic to the letter $C$?

Which letter is not homeomorphic to the letter $C$?

I think letter $O$ and $o$ are not homeomorphic to the letter $C$. Is that correct?

Is there any other letter?

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From the wikipage on topology, here are the equivalence classes (under homeomorphism) of letters of the English alphabet. Any letter not in the same bracketed set as 'C' is not homeomorphic to it.

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Whether K is homeomorphic to H or X depends on the type being used. – Michael Greinecker Nov 1 '12 at 15:47
Of course; that is the case for all letters. – Benjamin Dickman Nov 1 '12 at 15:51
Of course. It's just more subtle here. I was staring for some at the K, trying to think of a homemorphism until I realized the very subtle space between two crossing. – Michael Greinecker Nov 1 '12 at 15:53
Note also that 'upper' is the case for all letters. – Benjamin Dickman Nov 1 '12 at 15:54
@B.: what about D, O? – tomasz Nov 1 '12 at 16:18

There are many others $E$ or $Q$ for example. The most basic method I know of is by assuming there is one, then it restricts to the subspace if you take out one (or more) points. Then the number of connected components of this subspace is an invariant.

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There's a definitional question here: $E$ is homeomorphic to $C$ if you think of letters as two-dimensional blobs, but not if you think of them as graphs. – Micah Nov 1 '12 at 15:43

O is not homeomorphic to C: they're not even homotopic since O has one hole and C has zero. B has two holes and is homeomorphic to neither C nor O.

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That's right. Also, $B$ is not homeomorphic to $C$ since homeomorphic letters have same fundamental groups but $\pi_1(C) = 0$ and $\pi_1(B) = \mathbb Z \ast \mathbb Z$.

Alternatively, you could argue that if you remove any point from $C$, except one of the two endpoints, then $C$ is no longer connected whereas if you remove one point from $B$, no matter which point you remove, $B$ will still be a connected space.

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Hope this helps. Now try to compile a full list of all letters that aren't homeomorphic to $C$, including the argument why they aren't. – Rudy the Reindeer Nov 1 '12 at 15:39
I think usually '$\times$' denotes direct product and '$*$' denotes free product. Following this notation, $\pi_1(B)=\mathbb{Z}*\mathbb{Z}$. – 23rd Nov 1 '12 at 15:50
Dear @richard Thank you for pointing out this typo! – Rudy the Reindeer Nov 1 '12 at 16:16
@MattN.: I think the argument by number of points you need to remove to disconnect the space is more elementary, at least if you consider zero-width letters... – tomasz Nov 1 '12 at 16:21
@MattN.: Touche. Rather, you can look at the number of points you can remove without disconnecting the space. ;) – tomasz Nov 1 '12 at 16:36