Surfaces of Finite Topological Type When I read some papers, I see the term 'Surfaces of Finite Topological Type'. But it is not defined. I guess that it is used for surfaces with finite Euler characteristic. But I could not be sure.
 A: A surface $S$ has finite type if there exists a compact surface $F$ and a finite subset $A \subset F$ such that $S$ is homeomorphic to $F-A$. 
To speak about "finite Euler characteristic" is somewhat fishy: the Euler characteristic is not even well-defined unless some other quantities are finite. At a minimum, you need all of the homology groups $H_n(S;\mathbb{Z})$ to be finitely generated in order to even define the Euler characteristic (namely as the alternating sum of the ranks of those homology groups).
However, it is true that a surface $S$ has finite type if and only if all of its homology groups $H_n(S;\mathbb{Z})$ have finite rank. The trouble with this statement is that it is hard to prove the "if" direction.
A: From W.H. Fleming's Nondegenerate Surfaces of Finite Topological Type, the topological type of a surface $M$ (I've taken the liberty to modernize notation here) with Euler characteristic $\chi(M)$ and $v$ punctures, the topological type of $M$ is $$ \tau =(\chi(M), v , \epsilon)$$ 
where $\epsilon$ denotes whether $M$ is orientable or no. Further $M$ is said to be of finite topological type iff it is of the type of a compact 2-manifold.
A: An open (non-compact and complete) manifold M is said to have finite topological type, if it is homeomorphic to the interior of a compact manifold with boundary. Recall that according to Cheeger-Gromoll’s soul theorem, an open manifold with non-negative sectional curvature always has finite topological type, which is false if one relaxes the assumption on sectional curvature to Ricci curvature for the case of dimension ≥ 4.
