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I have what seems like a very silly argument, but I can't figure out what's wrong with it.

Suppose that I'm trying to calculate $H_1$ of a solid torus $X$ (a torus with its ``interior'' filled in). On the one hand, it's homotopy equivalent to a circle, so $H_1(X) \cong \mathbb Z$.

On the other hand, $H_k(X)$ is isomorphic to $H_k$ of the $(k+1)$-skeleton of $X$, and in this case (letting $k = 1$) that would just be the regular torus; but then $H_1(X) = H_1(T^2) \cong \mathbb Z^2$.

What's going on?

fish
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1 Answers1

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The $2$-skeleton of a filled torus is not a torus (I say "the", but there are many ways to realise a torus as a CW-complex, so "a" is more correct). You can't attach the 3-cell (the "fill") to the hollow inside of a regular torus, as that void isn't homeomorphic to an open ball.

One possible 2-skeleton of the filled torus is a hollow cylinder with top and bottom, with the top and bottom glued together. This top/bottom cell kills one of the copies of $\Bbb Z$ in the homology.

Arthur
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  • How does this gluing happen? Is a hollow torus with top and bottom identified not the canonical description of a hollow torus? – semper-lux Aug 26 '22 at 19:20
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    @Migillope The top and bottom of the cylinder are (filled) discs, not circles. The cylinder is a closed container, not an open pipe. After gluing, what you have is a hollow torus with one disc blocking the tube. This is necessary as the 3-cell of the filled torus needs to fill a simply-connected void, and the standard (hollow) torus has no such thing. – Arthur Aug 26 '22 at 19:25
  • Thank you, that clears it up! – semper-lux Aug 27 '22 at 04:45