# Homology group of space $X$ given by the labelling scheme $aabcb^{-1}c^{-1}$

I have to calculate the homology group of the quotient space $X$ given by the labelling scheme $aabcb^{-1}c^{-1}$ and then determine to which of the following spaces it is homeomorphic: $S^2, P_1,P_2,\dots, T_1,T_2,\dots$ where $P_m$ is $m$-fold connected sum of projective planes and $T_m$ is the $m$-fold connected sum of tori. The definition is $H_1(X)=\pi_1(X,x_0)/[\pi_1(X,x_0),\pi_1(X,x_0)]$. In Munkres there is the following corollary which I want to use:

Corollary 75.2: Let $F$ be a free group with free generators $\alpha_1,\dots,\alpha_n$. Let $N$ be the least normal subgroup of $F$ containing the element $x$ of $F$. Let $G=F/N$. Let $p\colon F\rightarrow F/[F,F]$ be projection. Then $G/[G,G]$ is isomorphic to the quotient of $F/[F,F]$, which is free abelian with basis $p(\alpha_1),\dots,p(\alpha_n)$, by the subgroup generated by $p(x)$.

In my case, we have $F$ is the free group with free generators $a,b,c$, N is the least normal subgroup of $F$ containing the element $a^2bcb^{-1}c^{-1}$, $G$ is the fundamental group and $G/[G,G]$ is the homology group. The corollary says that the homology group is isomorphic to $F/[F,F]$ divided by the least normal subgroup containing the element $q(a^2bcb^{-1}c^{-1})=a^2$ (call this normal subgroup M) because $F/[F,F]$ is abelian. Since $F/[F,F]$ is free abelian of rank 3, it is isomorphic to $\mathbb{Z}\times \mathbb{Z}\times\mathbb{Z}$. So dividing this by $M$ we have that the homology group is isomorphic to $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}$.

Now, the fundamental group of $S^2$ is trivial, so $X$ is not homeomorphic to $S^2$. The labelling scheme for the $m$-fold sum of projective planes $P_m$ is given by $(a_1a_1)(a_2a_2)\dots (a_ma_m)$ so in the same fashion the homology group should be the $m$-fold product of $\mathbb{Z}/2\mathbb{Z}$ with it self. The homology group of the $n$-fold connected sum of tori $T_n$ is free abelian of rank $2n$ because its labelling scheme is given by $[a_1,b_1]\cdots[a_n,b_n]$ where $[a_i,b_i]$ is the commutator.

Now I don't understand to which of these spaces $X$ is homeomorphic to, because its homology group is not isomorphic to one of the mentioned spaces. What did I wrong here? I need help. Thanks.

The space is isomorphic to $P_3$, using your notation. There are a few ways to show it, including by a cut-and-paste type argument (draw the fundamental regions for each of these, cut them and glue them back together, labeling the edges appropriately to turn one into the other...).
One way to see it is to note that $P_3 \cong T_1 \sharp P_1$. From this it is easy to see that the first homology group of $P_3$ is exactly $\mathbb{Z}/2 \oplus \mathbb{Z} \oplus \mathbb{Z}$, which is exactly what you have. In general, it is worth noting that $P_{2n+1} \cong T_n \sharp P_1$. However, to show this requires either a classification argument or an explicit cut-and-paste argument, which may not be easier.
• The labelling scheme for $P_3$ is $aabbcc$. So if I calculate the homology group as I did, I get $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$, because I have to divide the free abelian group of rank 3 by the normal subgroup containing the element $q(a^2b^2c^2)=a^2b^2c^2$. Is this calculation wrong? – Badshah Aug 29 '14 at 9:30
• That isn't quite correct. Remember that in $H_1(X)$, you abelianize the fundamental group. So when we abelianize the group in question, we get the single relation $2a + 2b + 2c = 0$---not $2a = 2b = 2c = 0$! If it helps, consider the two possible presentations for the fundamental group of the Klein bottle: $a^2b^2 = 1$ and $abab^{-1} = 1$, which are basically analogous. – Simon Rose Aug 29 '14 at 11:54
• To be clear, what you end up with is an abelian group on three generators, $a, b, c$, which satisfies the single relation $2a + 2b + 2c = 0$. So if you then choose a new basis for this group given by $a, b, a + b + c$, you can hopefully see the isomorphism in question. – Simon Rose Aug 29 '14 at 11:56