# A question regarding the notation/definition of homology group

I'm confused about the different types of homology groups. For instance, on Wikipedia (Orientability) they write $H_n (M, \partial M; \mathbb{Z} )$.

On the other hand, where they explain the construction of homology groups, they write $H_n (X) := ker(\partial_n) / im(\partial_{n + 1})$. (here)

Are these two the same?

The reason why I'm interested in this question is because I'm interested in an alternative definition of orientability, namely, that a compact $n$-manifold is orientable if and only if its top homology group is isomorphic to $\mathbb{Z}$. In fact, I'm only interested in surfaces, i.e. $n= 2$.

Seeing as every surface can be triangulated, I assume I can restrict myself to looking at simplicial homology groups. Are there any further restrictions that I can make to make it easier?

Thanks for any help.

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The notation $H_n(X,A;R)$ for $A\subset X$ denotes relative homology with coefficients in the ring $R$. This is a more general construction that gives the usual homology $H_n(X;R)$ when $A=\emptyset$. –  Andrea Mori Mar 2 '11 at 11:55

If $S$ is your surface, you should consider $H_2(S,\partial S;Z)$, which means the second homology relative to the boundary, with $Z$ coefficients. The notation is used for many different kinds of homology, but you are right, it is fine to use the simplicial version. The default coefficients are (often) $Z$ so sometimes they are left out of the notation. Thus if the surface is closed, you may just see $H_2(S)$ and if the surface has boundary you may see $H_2(S,\partial S)$ (without the $Z$'s).
The coefficients, while sometimes omitted, are important. For instance if you use $Z/2Z$ coefficients then the second homology of a closed surface will be $Z/2Z$, whether it is orientable or not.
Anyway, with $Z$ coefficients, if $S$ has no boundary then $H_2(S)$ will be $Z$ if $S$ is orientable and $0$ if $S$ is non-orientable. If $S$ has boundary then $H_2(S,\partial S)$ will be $Z$ if $S$ is orientable and $0$ if $S$ is non-orientable.