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.
Consider the embedding $\psi:\Bbb R\times\Bbb R\big/{\Bbb Z\times\Bbb Z}\hookrightarrow\Bbb R^3$
$$\psi([\theta,\tau])=
\begin{pmatrix}
\cos(2\pi\theta) &-\sin(2\pi\theta) & 0\\
\sin(2\pi\theta) &-\sin(2\pi\theta) & 0\\
0&0&1
\end{pmatrix}
\begin{pmatrix}
2+\cos(2\pi\tau)\\
\sin(2\pi\tau)\\
0
\end{pmatrix}$$
Then one verifies
(on the picture and then by doing the math) that $-\psi([\theta,\tau])=\psi([\theta+\frac12,-\tau])$
Best Answer
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.