Difference between Euler characteristics of a Riemann surfaces Let $X$ be a compact connected Riemann surface of genus $g$. Let $U$ be the complement of $r$ points in $X$. 
The Euler characteristic of $X$ = $2-2g$. That I understand. But I'm confused about the difference between
the Euler characteristic of $U$ and the Euler characteristic of $U$ with compact support.
Are they equal?
I suspect one of them equals $2-2g-r$, but which one?

The usual Euler characteristic is a homotopy invariant so you can calculate it for $U$ by replacing $U$ with something homotopy equivalent. I believe $U$ deformation retracts to a wedge sum of $2g + r - 1$ circles. This computes the Euler characteristic as $1 - (2g + r -1) = 2 - 2g - r$ using the fact that the homology of a wedge sum is the direct sum of hommologies. 
On the other hand, Euler characteristic with compact supports satisfies cut-and-paste so that if $U = X \setminus S$ where $S$ is a finite set of $r$ points, $\chi_c(U) = \chi_c(X) - \chi_c(S) = 2 - 2g - r$ where we used the fact that compactly supported Euler characteristic is the same as the usual Euler characteristic for compact spaces.  
So in this case you seem to get the same answer either way. This isn't true in general. However, I think in this case we could have argued that $\chi(U) = \chi_c(U)$ without calculating it as follows. Take a small open disc around each point of $S$ so that the discs are disjoint. Call the union of these discs $V$. Then $V \cup U = X$ and $V$ and $U$ form an excisive pair and so $\chi$ satisfies cut-and-past for this cover. 
In particular, $\chi(X) = \chi(U) + \chi(V) - \chi(U\cap V)$ but $U\cap V$ is a disjoint union of annuli which are homotopy equivalent to circles and have Euler characteristic $0$ and $V$ is a disjoint union of discs which deformation retract to their centers. Thus $\chi(X) = \chi(U) + \chi(S)$ and we arrive at the same cut and paste formula that compactly supported Euler characteristic satisfies. 
