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From Harvard math qualification exam, 1990.

Let $X$ be a smooth manifold with an open cover $N<\infty$ sets $\{B_{n}\}^{N}_{1}$ which are contractible. Assume that $$\pi_{0}(B_{n}\cap B_{m})\le k, \forall n,m$$ for all $n$ and $m$. Give an upper bound to the first betti number of $X$.

Recall that the first betti number is defined to be the rank of $H_{1}(X)$. The main trouble is how the conditon on $\pi_{0}$ relates to the bound. My thought (perhaps dumb) is to proceed this inductively. But an arbitrally large genus surface can be covered by a large enough open ball that is connected. So in this case $n=2g, k=0, N=1$.

I asked my professor today, and he gave the hint that this is nothing but Cech cohomology, in which the $B_{i}$s resemble the faces in simplicial homology. He suggest since $X$ has such a cover then it is the same as putting a simplicial structure on $X$. So the first betti number must be bounded. But how large is the bound given $N$ and $k$? Imagine a two hole torus covered by a million small open balls with one next to each other, then $k$ should be 2 if the balls are small enough. But this seems to carry over to a 3-hole torus or a torus of any genus. I suspect there is some fundamental misunderstanding on the statement of the problem or concepts behind, so I venture to ask. A closer look at the Cech cohomology article does not help me to come up with a bound.

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    Couldn't you try an induction and Mayer Vietoris? I have not thought through it, but it might be worth thinking about...2012-08-07
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    As I wrote in the question, I do not know how to handle the base case.2012-08-07
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    Well if $N=1$ then your manifold is contractible and hence has zero as first betti number.2012-08-07
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    I don't think the bound is supposed to be a _good_ one, just a finite one!2012-08-07
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    yeah, I think this way too! Just not sure if I am confused.2012-08-07
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    Mayer-Vietoris and induction looks good to me. For $N = 1$, you get $b_1 = 0$ because $M = B_1$, which is contractible (as mland says). Each time you add a $B_i$ you add at most $k-1$ (the maximum rank of $H_0(B_i \cap (B_1 \cup \ldots \cup B_{i-1}), *)$) to $b_1$. I get an upper bound of $(N-1)(k-1)$ (which is sharp for $M = S^1$ covered by two open intervals so that $N = k = 2$ and $b_1 = 1$).2012-08-07
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    No, you can cover a $n$ hole torus (a closed surface) by a large enough open ball. It still works as an open cover and is contractible, but this give no bound on the genus of the torus.2012-08-07
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    The sets in the open over here are intended to be open subsets of the manifold. I think you are thinking of the torus as embedded in Euclidean space, but you should be thinking of the manifold as a topological space in its own right. THe only open covering by one set is the whole torus and that isn't contractible.2012-08-08
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    This clarified my main confusion in the problem. I was thinking about a $n$-genus surface embedded isometrically in $R^{3}$ using Nash's theorem. I agree thinking intrinsically is more appropriate.2012-08-08
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    On reflection, my estimate was wrong, because I underestimated the number of connected components of $B_i \cap (B_1 \cup \ldots \cup B_{i-1})$ (which in a worst case could be $ik$). With the correction, the Mayer-Vietoris method comes up with the same answer as Zhen Lin (but without the sheaves!).2012-08-08
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    sheaves are not that scary, I guess I have to work everything myself from scratch once to appreciate the problem. I do not want to be lazy...2012-08-08
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    I see your computation. Assume $B_{i}$ are open subsets of $X$, then we should have $0\rightarrow H_{1}(X)\rightarrow H_{0}(B_{i}\cap (B_{1}\cup...\cup B_{i-1} ))\rightarrow 0$. The rest just follows by standard induction. Since every time we add $B_{i}$ as you noted the worst case is $ik$, $(\sum^{N}_{1}i)k=\frac{N(N-1)}{2}k$. Thanks!2012-08-08

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If $\mathfrak{U}$ is any open cover of $X$, the Čech-to-derived functor spectral sequence assures us of an exact sequence of the form $$0 \longrightarrow H^1(\check{C}^\bullet(\mathfrak{U})) \longrightarrow H^1(X, \mathbb{Z}) \longrightarrow H^0(\check{C}^\bullet(\mathfrak{U}, \mathscr{H}^1(\mathbb{Z}))) \longrightarrow \cdots$$ where $\mathscr{H}^1(\mathbb{Z})$ is the presheaf $U \mapsto H^1(U, \mathbb{Z})$. (Here, $H^1$ refers to sheaf cohomology, but this is the same as singular cohomology when $X$ admits a triangulation.) Since we are assuming the open sets of $\mathfrak{U}$ are contractible, we have $\check{C}^0(\mathfrak{U}, \mathscr{H}^1(\mathbb{Z})) = 0$. So in fact we have an isomorphism $$H^1(\check{C}^\bullet(\mathfrak{U})) \cong H^1(X, \mathbb{Z})$$ and the universal coefficient theorem gives a short exact sequence $$0 \longrightarrow \textrm{Ext}^1(H_0(X, \mathbb{Z}), \mathbb{Z}) \longrightarrow H^1(X, \mathbb{Z}) \longrightarrow \textrm{Hom}(H_1(X, \mathbb{Z}), \mathbb{Z}) \longrightarrow 0$$ but if $X$ is path-connected then $H_0(X, \mathbb{Z})$ is a finitely-generated free abelian group, so $\textrm{Ext}^1(H_0(X, \mathbb{Z}), \mathbb{Z}) = 0$, so there is an isomorphism $$H^1(X, \mathbb{Z}) \cong \textrm{Hom}(H_1(X, \mathbb{Z}), \mathbb{Z})$$ and therefore the ranks of $H^1(X, \mathbb{Z})$ and $H_1(X, \mathbb{Z})$ are equal when finite. Thus, $$b_1(X) = \operatorname{rank} H^1(\check{C}^\bullet(\mathfrak{U}))$$ By simple linear algebra, $\operatorname{rank} H^1(\check{C}^\bullet(\mathfrak{U})) \le \operatorname{rank} \check{C}^1(\mathfrak{U})$, and $$\operatorname{rank} \check{C}^1(\mathfrak{U}) = \sum_{\substack{\{U, V\} \\ U \ne V}} b_0 (U \cap V) \le \frac{1}{2} (N-1) N k$$ by definition of $\check{C}^1(\mathfrak{U})$.

In summary, if $X$ admits a cover by $N$ contractible open subsets whose pairwise intersections have at most $k$ connected components, then: $$b_1(X) \le \frac{1}{2} (N-1) N k$$

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    Thanks! I need sometime to understand all you had written, but how could the bound be independent of $k$?2012-08-07
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    @user32240 Oops, I accidentally assumed $k = 1$. I've fixed my answer.2012-08-07