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2-D CW-complex.

If Z is a CW complex with one 0-cell.

For simplicity, the $\pi_1(X)$ has presentation where the generators are the 1-cells and the relation come for the 2-cells. More precisely, each 1-cell is a loop in CW complex. The c stands for closure finite and w for weak topology.

A CW complex can be defined inductively. A 2-D ddimensional CW complex is just a set of 2...

Can you please tell if what I did is right, or not.. If wrong what to use and how to solve?

Sam Hoffman
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dismay
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1 Answers1

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Let $a,b$ denote the two circles in $S^1 \vee S^1$. Your group $\mathbb{Z} \oplus \mathbb{Z}/2$ has a presentation $$ \mathbb{Z} \oplus \mathbb{Z}/2 = \langle a,b \mid a^2,\; aba^{-1}b^{-1} \rangle. $$ As in the proof of corollary 1.28 in Hatcher, we construct a space $X$ as the pushout of the following commutative diagram:

$\hspace{4cm}$

In words, this says that $X$ is the space obtained by gluing two disks $D^2$ to the space $S^1 \vee S^1$ via the maps $a^2$ and $aba^{-1}b^{-1}$. Of course it's hard to actually visualize this space, but at least it's clear how the $D^2$'s are glued to the loops of $S^1\vee S^1$. The space $X$ is clearly a $2$-dimensional path-connected CW-complex, since we glued only $2$-cells to $S^1 \vee S^1$. Then, by a common fact (or Hatcher proposition 1.26), we get $$ \pi_1(X) \cong \pi_1(S^1 \vee S^1) / \langle a^2,\; aba^{-1}b^{-1} \rangle = \langle a,b \mid a^2,\; aba^{-1}b^{-1} \rangle = \mathbb{Z} \oplus \mathbb{Z}/2, $$ so we have found a $2$-dimensional CW-complex which has the correct fundamental group. I hope this helps!

jasnee
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