# Rank of first homology group for surface with punctures?

I feel like this question will be a head-slapper once I figure out the answer, but for the moment I'm having trouble!

Let $M$ be a compact, connected, orientable 2-manifold of genus $g$ with $b$ boundary components. The Euler characteristic $\chi$ can be expressed as

$$\chi = 2 - 2g - b$$

and also as

$$\chi = H_0 - H_1 + H_2$$

where $H_i$ is the rank of the $i$th homology group. Since $M$ is connected we have $H_0 = 1$; since $M$ is a 2-manifold we also have $H_2=1$ (here I'm thinking about de Rham cohomology: $(H_2 = H^0 = \mathrm{ker}\ d_0/\mathrm{im}\ d_{-1} = \mathrm{ker}\ d_0$, which is just the constant functions.). Solving for $H_1$ yields

$$H_1 = 2g + b.$$

Sounds ok for the most part -- for instance, if $b=0$ then there are twice as many basis loops as handles. But what if, say, $g=0$ and $b=1$, i.e., a disk? Then we have $H_1 = 1$. But every loop in a disk is contractible! Help! (Seems like it should be something like $H_1 = 2g + b-1$ when $M$ has boundary and $H_1 = 2g$ otherwise, but I don't understand why from the arguments above.)

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Well, yes, the $n$-th homology group of an $n$-dimensional compact oriented manifold without boundary is $\mathbb{Z}$. – M.B. Jun 1 '12 at 22:05

The problem is your assumption that $\dim H_2 = 1$ (it is confusing to use the same symbol to denote a thing and the dimension of that thing). In fact the disk has trivial $H_2$ because it's contractible. Poincaré duality, as M.B. says in the comments, holds for (connected) compact oriented manifolds without boundary, so doesn't apply when $b > 0$.
I think I get it now -- forgetting about cohomology, I can just think of the second homology group as $H_2 = \mathrm{ker}\ \partial_2 / \mathrm{im}\ \partial_3 = \mathrm{ker}\ \partial_2$, i.e., I'm essentially looking for 2-dimensional subsets $A \subset M$ without boundary. Well, for a surface with boundary there are no such subsets so $H_2=0$; for a surface without boundary I have just one ($A = M$), so $H_2=1$. From there everything works out. – fuzzytron Jun 1 '12 at 22:23
Gotcha -- so that explains why $H_{dR}^0$ doesn't lead to the right answer. Thanks! – fuzzytron Jun 1 '12 at 22:24