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.
Fundamental group of an orientable surface of infinite genus.
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algebraic-topology
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0Can 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. – 2012-12-09
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0Well,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. – 2012-12-09
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3There'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. – 2012-12-09
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0can 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? – 2013-03-25
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0@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! – 2014-10-22
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1I was describing homotopy-type, not homeomorphism type. – 2014-10-22