general-topologyklein-bottlemanifolds
I have read in several places that the Klein bottle is a 2-manifold, but I cannot find an explicit proof anywhere.
How would you show it is locally Euclidean, Hausdorff and second countable (I think this is what defines something as a manifold…)?
And why can it not be embedded in $\mathbb R^3$?
Any help would be greatly appreciated.
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.
We can consider the Klein Bottle $K$ to be the quotient space of the cylinder $C = S^1 \times I$ by the identification $(z,0) \sim (z^{-1},1)$. Here we consider $S^1$ as the unit circle in the complex plane. Let $L = 1 \times I$. I claim that $K$ retracts onto the loop $L'$ defined by $L$ under the quotient map.
Let $q : C \to K$ be the quotient map. The map $q$ satisfies the following universal property:
If $g : C \to Z$ is a continuous map from $C$ to any topological space $Z$ such that $a \sim b$ implies $g(a) = g(b)$ for all $a,b \in C$ then there exists a unique map $f : K \to Z$ such that $g = f \circ q$.
Define a map $g : C \to S^1$ as follows: First project onto $L$, then quotient to $L'$. This is a continuous map which is preserved under $\sim$, hence a continuous map $f : K \to L'$ exists and must be the identity on $L'$ by the universal property.
Best Answer
The results you need in order to prove that the Klein bottle can't be embedded in $\mathbb{R^3}$:
$\bullet$ The Klein bottle isn't orientable.
$\bullet$ The Klein bottle is a compact, and connected 2-manifold.
$\bullet$ Any compact, connected manifold without boundary of dimension $n$ embedded in $\mathbb{R^{n+1}}$ is orientable.
Therefore, the Klein bottle can't be embedded in $\mathbb{R^3}$.