# Topology on Klein bottle?

I was trying to show that the Klein bottle was second countable. My try was to use that it has the subspace topology of $\mathbb R^3$. Then I noticed that it is not imbeddable into $\mathbb R^3$. Therefore one cannot use the subspace topology. But I don't know what else to do.

How to show that the Klein bottle is second countable? If it does not have subspace topology of $\mathbb R^3$, what topology does it have? Does immersions induce a topology?

Therefore: what is the topology on Klein bottle if Klein bottle is square with sides identified?

Thank you for help.

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You can embed it in $\mathbb R^4$ though... –  JSchlather Feb 4 '13 at 8:55
A Klein bottle is homeomorphic to a square with opposite edges identified in a particular way. –  Gerry Myerson Feb 4 '13 at 9:01
How do you define a Klein bottle? If you use @Gerry's homeomorphism as its definition (as on Wikipedia), then relatively little work is needed. –  Willie Wong Feb 4 '13 at 9:18
@goobie you can define a subspace/subset topology for any $D\subset X$: $O\subset D\quad \text{open} \Leftrightarrow \exists O'\subset X \quad\text{open} \wedge O'\cap D=O$ –  CBenni Feb 4 '13 at 13:17

Of course you could use the embedding of the Klein bottle $K:=I^2/R$ in $\mathbb R^4$, where $R$ is the equivalence relation. But you would need the exact formula for the mapping which can be fairly complicated.
You can also do it directly with the quotient space itself. The quotient map $q$ is surjective and continuous (by definition of the quotient space), but it is an easy exercise to show that $q$ is also closed, i.e. the preimage of the image of a closed set in $I^2$ is closed. Furthermore the fibers (the preimages of points) are the equivalence classes, and these are finite, thus compact. It follows that $q$ is a so called perfect mapping. The good thing about perfect maps is that they carry over many properties to the image, among which is second-countability.

Edit: (the OP asked for a hint regarding the closedness of the quotient map)

Let $C$ be closed in $I^2$ and let $\hat C$ its saturation, i.e. $\hat C=q^{-1}(q(C))$. You have to show that a point $x$ outside $\hat C$ is contained in an $\epsilon$-ball disjoint from $\hat C$. We will show this for $x$ being the upper right corner of the square. We have to find $\epsilon>0$ such that $B_\epsilon(x)\cap \hat C=\emptyset$, which is equivalent to $\hat B_\epsilon(x)\cap C=\emptyset$. The saturation of the ball will look like the red area in the figure below. Since the points in the upper right, the upper left, and the lower left corner are outside of $C$, it is obvious that such an $\epsilon$ exists. The cases when $x$ is a point on the edge or in the interior of the square are equally easy.
This shows that $q(C)$ is closed since its preimage $\hat C$ is.

Edit 2: (explicit formula for countable base of K)

Let $(B_k)_{k\in\mathbb N}$ be the countable base for $I^2$. Define $\mathscr B=\{K\backslash q(I^2\backslash(B_{k_1}\cup B_{k_2}\cup B_{k_3}\cup B_{k_4}))\ ;\ k_i\in\mathbb N\}$. Try to show that this constitutes a base of $K$. If you need help, feel free to ask.

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I see. Is there no way to show directly? I tried to write down an explicit countable basis by taking rational open balls in square and add the necessary balls at the boundary of the square. But then I failed to show it's a basis. –  goobie Feb 5 '13 at 9:46
Also I fail to prove that if $C$ is closed in $I^2$ then $\pi(C)$ is closed in $I^2 / \sim$. Could you give me a hint? Thank you. –  goobie Feb 5 '13 at 10:42
Did bounty work and you get 50 points? It should award to accepted answer. –  goobie Mar 11 '13 at 14:46
No, I didn't get the points. Read the section about bounties in the FAQ. It is awarded to the best answer written after the bounty started, or an answer you accept during bounty period. So you should have manually awarded the bounty right at the beginning, or re-accepted my answer. Didn't you get notified when the bounty ended about what to do? Well, I didn't know about this rule either. I noticed the bounty and thought it would be awarded to me automatically. I guess you thought the same. –  Stefan Hamcke Mar 11 '13 at 15:15
Okay, these things happen :-) Maybe, sometime in the future, when you will have gathered more reputation, you could start a new bounty, but I don't urge you to do this now. –  Stefan Hamcke Mar 11 '13 at 15:23

I take it you understand the topology on the square.

The topology on the square with opposite sides identified is the quotient topology. The idea of the quotient topology is this: if $X$ is a topological space, and $R$ is an equivalence relation on $X$, then the quotient topology on $X/R$ (which is the set of equivalence classes of $X$ under $R$) is chosen to make the natural surjection $f:X\to X/R$ continuous. So, a subset of $X/R$ is open in the quotient topology if and only if its inverse image under $f$ is open in $X$.

In our case, $X$ is the square, and two points are equivalent if they get identified when you identify the pairs of opposite edges.

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Can I use the definition of quotient topology somehow to show Klein bottle is second countable? –  goobie Feb 5 '13 at 13:01
I'm sure you can. If you can show the topology on a square in ${\bf R}^2$ is second countable, and you understand how the quotient topology on the Klein bottle relates to the topology on the square, it shouldn't be that hard. –  Gerry Myerson Feb 5 '13 at 22:33

The Klein bottle $K$ is an interesting nonorientable surface. It can be presented as a quotient of ${\mathbb R}^2$ by some discrete group, as explained in other answers to this question. But "second countability" is not for a split of a second a question in connection with $K$. Any open set in $K$ is a union of two-dimensional disks with rational radius and rational coordinates of their center.

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I can't make any sense of that last sentence unless I take the Klein bottle to be a subset of ${\bf R}^2$. –  Gerry Myerson Feb 5 '13 at 22:36
@Gerry Myerson: When $[0,1]^2$ is a fundamental domain for $K$ take disks of the described kind with radii $<{1\over3}$. These project to bona fide euclidean disks on $K$. $-$ By the way, my answer is about the same as yours. I just wanted to advise the OP not to fuss about second countability. –  Christian Blatter Feb 6 '13 at 9:27