A simple example calculating Čech comohomology

Question

I'm currently reading "Differential Forms in Algebraic Topology" by Bott & Tu. Unfortunately, I do not have much backgrounds in this field, so my question might be due to missing some very basic concepts.

In chapter 13 Monodromy, the author proves two theorems :

1. If $\pi_{1}(N(\mathfrak{U}))$, the fundamental group of the nerve of a good cover, is equal to zero, then every locally constant presheaf on $\mathfrak{U}$ is constant.
2. For a topological space $X$ with good cover $\mathfrak{U}$, we have $\pi_{1}(X)\simeq \pi_{1}(N(\mathfrak{U}))$.

I did have a difficult time understanding the proofs of the two theorems (not quite sure that I actually did understand them), but what really frustrates me are the examples and exercises at the end of the section:

Example 13.5

Let $S^{1}$ be the unit circle in the complex plain with good cover $\mathfrak{U}=\{U_{0}, U_{1},U_{3}\}$ as in the picture below.

The map $\pi : z \rightarrow z^2$ defines a fiber bundle $\pi : S^1 \rightarrow S^1$ each of whose fibers consists of two distinct points. Let $F=\{A,B\}$ be the fiber above the point 1.

The cohomology $H^*(F)$ consists of all functions on $\{A,B\}$, i.e., $H^*(F)=\{(a,b)\in \mathbb{R}^2\}$ . (I don't understand this part. Why is this true? Could someone explain?)

The author proceeds and mentions that the presheaf $\mathcal{H}^*(U)=H^*(\pi^{-1}U)$ is not a constant presheaf.

Exercis 13.6 tells me to compute the Čech cohomology $H^*(\mathfrak{U},\mathcal{H}^0)$ directly, noting that $H^*(S^1)=H^*_D\{C^*(\pi^{-1}\mathfrak{U},\Omega^*)\}=H_{\delta}H_d=H^*(\mathfrak{U},\mathcal{H}^0)$.

By using "indirect" methods, I would get $H^0(\mathfrak{U},\mathcal{H}^0)=H^1(\mathfrak{U},\mathcal{H}^0)=\mathbb{R}$ from above, right?

But I'm having trouble calculating the cohomology directly. In order to compute, I think I'd need to know:

1. what $C^0(\mathfrak{U},\mathcal{H}^0)$ and $C^1(\mathfrak{U},\mathcal{H}^0)$ looks like.
2. what the transition map $\rho$ 's looks like.
3. how the difference operator $\delta_0$ behaves
4. the image and kernel of $\delta_0$

From $H^*(F)=\{(a,b)\in \mathbb{R}\}$, I guess $C^0(\mathfrak{U},\mathcal{H}^0)$ and $C^1(\mathfrak{U},\mathcal{H}^0)$ would be something like $\mathbb{R}^2 \oplus \mathbb{R}^2\oplus \mathbb{R}^2$, but from then on, I'm completely lost. In addition, I don't see how this is related to monodromy dealt in this chapter.

I'm pretty sure that these questions are quite easy to answer, but could someone give me a in-depth explanation? I would like to solve the next two problems on my own!

Answer 1

As you said, $C^0(\mathfrak{U},\mathcal{H}^0)=\mathbb{R}^2\oplus\mathbb{R}^2\oplus\mathbb{R}^2$ and the same isomorphism holds for $C^1$. Now, we need to compute the differential, and for this, we need to understand the restriction map.

Recall that $\pi:\pi^{-1}(U)\to U$ is the map $z\mapsto z^2$. In other words, $\pi^{-1}(z)$ is the set of square roots of $z$. So if $U$ is an arc, $\pi^{-1}(U)$ is the union of two arcs and you can think of it as follows : if you pick $z\in U$, you have two choices of square roots, now when $z$ moves in $U$, the two choices moves continuously and forms two arcs.

Now let $z_0\in U_0\cap U_1, z_1\in U_1\cap U_2$ and $z_2\in U_2\cap U_0$. Write $z_i^a, z_i^b$ for the two choices of square roots of $z_i$. We won't make arbitrary choices for them. Instead, we will define them as follow :

• Define $z_0^a$ and $z_0^b$ arbitrarily (as the two square roots of $z_0$).
• When $z_0$ moves to $z_1$ in $U_1$, $z_0^a$ moves to a square roots of $z_1$. This will be $z_1^a$. And $z_0^b$ will move to $z_1^b$.
• When $z_1$ moves to $z_2$ in $U_2$, $z_1^a$ moves to $z_2^a$ and $z_1^b$ moves to $z_2^b$.

So now, you need to see the following :

When $z_2$ moves to $z_0$ in $U_0$ then $z_2^a$ moves to $z_0^b$ and $z_2^b$ moves to $z_0^a$.

Thus, using the identifications $\mathcal{H}^0(U_0)=maps(\{z_0^a,z_0^b\},\mathbb{R})$ and so on, we see that

• $\mathcal{H}^0(U_0)\to \mathcal{H}^0(U_0\cap U_1)$ is the identity
• $\mathcal{H}^0(U_1)\to \mathcal{H}^0(U_0\cap U_1)$ is the identity
• $\mathcal{H}^0(U_1)\to \mathcal{H}^0(U_1\cap U_2)$ is the identity
• $\mathcal{H}^0(U_2)\to \mathcal{H}^0(U_1\cap U_2)$ is the identity
• $\mathcal{H}^0(U_2)\to \mathcal{H}^0(U_0\cap U_2)$ is the identity
• $\mathcal{H}^0(U_0)\to \mathcal{H}^0(U_0\cap U_2)$ is the map $\tau:(a,b)\mapsto (b,a)$.

It follows that the operator $\delta_0$ is given by the block matrix (each block is a $2\times 2$ matrix) $$ \begin{pmatrix} Id & Id & 0\\ 0& -Id & Id\\ -\tau& 0& -Id \end{pmatrix}$$

Answer 2

It's funny because I was just about to post about this exact same cohomology, so I'll give you a partial answer that I've worked out (computation of $H^0$). I'm still confused about $H^1$. I would comment but I don't have the rep yet, so apologies for the partial answer.

I saw this problem in a video, and the background was to use the presheaf $F(U) = \mathbb{R}$ for non-empty $U$. I followed the definition of Cech cohomology from Wikipedia. The $0$-simplices are just the sets $\{U_0\}, \{U_1\}, \{U_2\}$. Similarly the $1$-simplices are the ordered sets $\sigma_{ab} = \{U_a, U_b\}$. Now, $H^0$ is just the kernel of $\delta_0$, which is the map $(\delta_0 f)(\sigma_{ab}) = f(\sigma_b) - f(\sigma_a)$. If $f$ is in the kernel, that means it is equal on each $0$-simplex. Hence, the kernel is the set of constant maps, which is of course equivalent to $\mathbb{R}$, since each cochain $f$ is mapping into $F(|\sigma|) = \mathbb{R}$.

Now, for $H^1$. It is the quotient of the kernel of $\delta_1$ and the image of $\delta_0$. $\delta_1$ maps into the set of $2$-cochains, but since the three open sets have an empty intersection, there are no $2$-simplices and hence no $2$-cochains. Then any $1$-cochain is in the kernel of $\delta_1$. Now, the image of $\delta_0$ are the $1$-cochains which satisfy $f(\sigma_{ab}) = h(\sigma_b) - h(\sigma_a)$ for some $0$-cochain $h$. Then the simplices $\sigma_{ab}$ and $\sigma_{ba}$ are no longer independent; $f(\sigma_{ab}) = -f(\sigma_{ba})$. In my simple mind, that would imply that $H^1 = \mathbb{R}^6 / \mathbb{R}^3 = \mathbb{R}^3$, but that's not quite right. Or maybe the argument is different. I don't know.