I'm going to use your Mayer-Vietoris idea for computing the integer cohomology of $T^1 K$, but with a twist.

To that end, let $\pi:T^1 K\rightarrow K$ be the natural projection map. I'll use $U'$ and $V'$ for the two Mobius band halves of $K$, which overlap in the boundary circle of each Mobius band. Let $U = \pi^{-1}(U')$ and $V = \pi^{-1}(V')$. Then $U$ and $V$ obviously form an open cover of $T^1 K$, so we can use them for Mayer-Vietoris. We just need to understand the topology of $U$, $V$, and $U\cap V$.

**Claim 1**: Both $U$ and $V$ deformation retract to copies of $K$.

For $U'$ deformation retracts to its core circle, and $\pi^{-1}$ of this deformation gives a deformtion of $U$ to $\pi^{-1}(S^1_c)$, with $S^1_c$ denoting the core circle in a Mobius band.

So, let's figure out what $\pi^{-1}(S^1_c)$ is. First, restricting the tangent bundle of $K$ to a Mobius band half gives the tangent bundle of the Mobius band. This restricts to a non-trivial $\mathbb{R}^2$ bundle over $S^1_c$ (just draw a picture!), so $\pi^{-1}(S^1_c)$ is a non-trivial $S^1$ bundle over $S^1$. That is, it's a Klein bottle as claimed. $\square$

**Claim 2:** The space $U\cap V$ deformation retracts to $T^2$.

The space $U'\cap V'$ deformation retracts to a circle winding around $S^1_c$ two times. I'll denote this by $S^1_d$ (where $d$ is for "double") Thus, $\pi^{-1}(U\cap V)$ deformation retracts to $\pi^{-1}(S^1_d)$. Then the tangent bundle of $K$ restrcits to the tangent bundle of $M$, which restricts to a *trivial* bundle over $S^1_d$ (again, just draw a picture). Thus, $\pi^{-1}(S^1_d)$ is a trivial $S^1$ bundle, so is $T^2$ as claimed. $\square$

**Claim 3:**: With respect to the deformations in Claim 1 and 2, the inclusion maps $U\cap V\rightarrow U,V$ are homotopic to the double covering $T^2\rightarrow K$.

For, if we imagine homotoping $S^1_d$ to the core circle $S^1_c$, we get a double cover map $S^1_d\rightarrow S^1_c$. And this double cover map acts as the identity on the $S^1$-fibers because coverings are local diffeomorphisms. $\square$

We can now compute $\pi_1(T^1 K)$ using Seifert-van Kampen. To that end, we'll write $\pi_1(T^2) = \langle a,b| aba^{-1}b^{-1}\rangle$, and $\pi_1(K) = \langle s,t| sts^{-1}t\rangle$. Then $\pi_1(K)$ has a unique index $2$ abelian subgroup, generated by $s^2$ and $t$, so we may assume the double cover $T^2\rightarrow K$ maps $a$ to $s^2$ and $b$ to $t$. Now, Seifert-van Kampen immediately gives:

**Claim 4:** We have $\pi_1(T^1 K)\cong \langle s,t, u,v| sts^{-1}t, uvu^{-1}v, s^2 u^{-2}, tv^{-1} \rangle$.

By the way, Claim 4 establishes that $\pi_1$ is *not* abelian, since there is a surjective map from $\pi_1$ to the order $8$ quaternion group $Q_8 = \{\pm 1, \pm i, \pm j, \pm k\}$ given by sending $(s,t,u,v)\mapsto (i,j,k,j)$. In particular, $T^1 K$ and $T^1 T^2$ are *not* homotopy equivalent.

Since $H_1$ is the abelianization of $\pi_1$, we can now compute $H_1$.

**Claim 5:** We have $H_1(T^1 K) \cong \mathbb{Z}\oplus (\mathbb{Z}/(2))^2$

For, if we assume all the variables commute, every element in $H_1(T^1 K)$ can be written in the form $s^\alpha t^\beta u^\gamma v^\delta$. But the relation $sts^{-1}t$ now implies that $t$ has order $2$, so we may assume $\beta \in \{0,1\}$. Likewise, $\delta \in\{0,1\}$. In addition, the relation $s^2u^{-2}$ means that we may assume $\gamma \in \{0,1\}$.

This gives a map from $H_1(T^1 K)$ to $\mathbb{Z}\oplus (\mathbb{Z}/(2))^2$, sending $s^\alpha t^\beta u^\gamma v^\delta$ to $(\alpha+\gamma,\beta + \delta,\gamma)$. We claim that this map is an isomorphism.

To see that it's well defined, we just need to check that each relation in $\pi_1(T^1K)$ is sent to the identity. But, e.g., $sts^{-1} t = t^2\mapsto (0,2,0) = (0,0,0)$, etc. Surjectivity is obvious. And injectivity is also easy to verify. $\square$

The calculation of $H_1$, together with your prior knowledge that $T^1 K$ is orientable, allows us to compute all homology groups.

**Claim 6:** The non-zero homology of $T^1 K$ is given by $$H_n(T^1 K)\cong \begin{cases}\mathbb{Z} & n=0, 2,3\\ \mathbb{Z}\oplus(\mathbb{Z}/(2))^2 & n=1 \end{cases}.$$

We obviously have $H_0(T^1 K) \cong \mathbb{Z}$, you've already noted that $H_3(T^1 K)\cong \mathbb{Z}$, and we just computed $H_1(T^1 K)$. So, all that remains is $H_2(T^1 K)$. We will compute this via a combination of Poincare duality and universal coefficients.

We start with the torsion subgroup of $H_2(T^1 K)$. By Poincare duality, this is the same as the torsion subgroup of $H^1(T^1 K)$. But by universal coefficients, $H^1(T^1 K)$ has torsion given by an Ext term involving $H_0(T^1 K)$. Since $H_0(T^1 K)$ is free, this Ext term vanishes, so $H^1(T^1 K)$ is torsion free. Thus, so is $H_2(T^1 K)$.

To compute the free part of $H_2(T^1 K)$, we proceed analogously. By universal coefficients, the free part of $H_2(T^1 K)$ is isomorphic to the free part of $H^2(T^1 K)$, which, via Poincare duality, is isomorphic to the free part of $H_1(T^1 K)$, which is isomorphic to $\mathbb{Z}$. $\square$.

## Best Answer

It seems like you might be confused regarding well-definedness of normal bundles and well-definedness of fiber dimension. So I will write an answer focussing on that issue.

The normal bundle of an $m$-dimensional manifold $M$ is not a well-defined thing. First you have to

choosean embedding (or an immersion) of $M$: only then can you talk about the normal bundle of $M$relative to that embedding (or immersion); and only then can you talk about the dimension of the fibers of that normal bundle.Maybe $M$ was already given to you as a submanifold of $\mathbb R^n$ so you can certainly talk about the normal bundle of the inclusion map (which is an embedding), but that should not deter you from studying other (noninclusion) embeddings of $M$ and their normal bundles.

For example, if you embed the circle $S^1$ into the plane $\mathbb R^2$ then the normal bundle of that embedding has fibers of dimension $2-1=1$, i.e. lines. But if on the other hand you embed $S^1$ as a knot in $\mathbb R^3$ then the normal bundle of this new embedding has fibers of dimension $3-1=2$, i.e. planes.

Regarding "

the" Klein bottle $K$ (scare quotes intentional), which is a 2-dimensional manifold, like any manifold it has many different embeddings. The lowest dimensional Euclidean space into which $K$ embeds is $\mathbb R^4$, and under any such embedding the normal bundle has fibers of dimension $4-2=2$. In fact you can embed $K$ into Euclidean space $\mathbb R^n$ of any dimension $n \ge 4$, and you would get a normal bundle with fibers of dimension $n-2$.The usual image of a Klein bottle in $\mathbb R^3$ is not an embedding but it

isan immersion, and one can still define the normal bundle of an immersion. And the same formula holds: the normal bundle of an $m$-dimensional manifold immersed into $\mathbb R^n$ has fibers of dimension $n-m$. So the normal bundle of a Klein bottle immersed into $\mathbb R^3$ has fibers of dimension $3-2=1$, i.e. lines.