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I am trying to calculate the fundamental group of an orientable surface $X$ of countably infinite genus. The $1$-skeleton $Y$ of $X$ is infinite wedge of circles, so its fundamental group is free group on countably infinite generators, but I am not able to see how is the $2$-cell attached to $Y$. My guess is that it is attached by loop of product of commutators of generators but this product being infinite doesn't make sense in group.

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Can you be more precise about which surface are you looking at? When I think of a surface of infinite genus, I imagine something like a torus extended indefinitely inside $\mathbb{R}^3$, with countably many holes. If this is what you have, are you sure that the $1$-skeleton is the infinite wedge of circles? In the infinite wedge of circles, they all have a point in common, which makes things weird when you pass to the surface. – student Dec 9 '12 at 6:44
Well,according to the exercise(1.16) in hatcher it is like an extended torus indefinitely inside $\mathbb R^3\$ ,with countably many holes but then I don’t think so it's 1-skeleton is wedge of infinite circles as such things doest sit inside euclidean space. – kuhu Dec 9 '12 at 6:49
There's no 2-cell. At least, not in the minimal CW-decomposition of the space up to homotopy-type. The space is homotopy-equivalent to a wedge of circles, full stop. – Ryan Budney Dec 9 '12 at 8:30
can someone please elaborate on why is the surface homotopy-equivalent to the wedge of circles please? Also, how could we see it by deformation retracting the surface onto a graph? – TJIF Mar 25 '13 at 1:10
@RyanBudney: Definitely there is a 2-cell. Since it is a surface, i.e. 2-dimensional manifold, it cannot be homeomorphic to a 1-dimensional CW-complex! – mathreader Oct 22 '14 at 4:36

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