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We know that $\langle a,b;(ab)^2=1\rangle$ and $\langle z;z^2\rangle$ are presentations of the fundamental group of the projective plane. Therefore, one is obtained from the other via Tietze transformations, but what I get is a contradiction, so I am doing something wrong and I don't know what it is: $$\langle a,b;(ab)^2=1\rangle\cong\langle a,b,c;(ab)^2=1,c=ab\rangle\cong\langle a,b,c;c^2=1,a=cb^{-1}\rangle \\\cong\langle b,c;c^2=1\rangle\cong\mathbb{Z}*\mathbb{Z}_2$$ which is not isomorphic to $\mathbb{Z}_2$. Where is the mistake?

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    $\begingroup$ Are you sure $<a,b;(ab)^2>$ is the presentation for the fundamental group of $RP^2$? $\endgroup$ – Ripan Saha Apr 21 '15 at 5:17
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    $\begingroup$ NB that the words $e, a, a^2, a^3, \ldots$ all represent different elements in $\langle a, b ; (ab)^2 = 1 \rangle$. $\endgroup$ – Travis Willse Apr 21 '15 at 5:27
  • $\begingroup$ Consider the square identifying the edges according the word $abab$. Draw a path connecting the starting points of the $a$ edges through the diagonal and name this path $c$. If we cut the square through $c$ and paste the remaining triangles identifying the $a$ edges, we get a square identifying the edges according the word $ccb^{-1}b$. Then identifying the $b$ edges we get the 2-polygon of edges $cc$. This is homeomorphic to the projective plane right? $\endgroup$ – Chilote Apr 21 '15 at 5:33
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    $\begingroup$ It is, but the way you're computing the fundamental group is not quite correct. Ignore the 2-cell, and pay attention to what happens with the edges. When you glue them together, you don't get two circles glued together at a point, like with the torus; you get a single circle, with the top hemisphere corresponding to $a$, and the bottom hemisphere corresponding to $b$. This has fundamental group $\langle ab \rangle$. When you glue on your 2-cell, you get fundamental group $\langle ab | (ab)^2 \rangle$. As you noticed here, what the 1-cells are doing before you look at the 2-cell matters a lot! $\endgroup$ – user98602 Apr 21 '15 at 5:46
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    $\begingroup$ @MikeMiller $\langle a,b\mid (ab)^2\rangle$ is neither a presentation of the $\pi_1$ of the projective plane, nor of the Klein bottle (the latter is torsion-free). It is not $\pi_1$ of any surface, even with boundary (the only surface with non-torsion-free $\pi_1$ is the projective plane). $\endgroup$ – YCor Apr 21 '15 at 10:27
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The use of Tietze transformations to prove that $\langle a,b;(ab)^2\rangle\cong \mathbb{Z}*\mathbb{Z}_2$ is correct. There are no mistakes in that.

The mistake is to claim that $\langle a,b;(ab)^2\rangle$ is a presentation of the projective plane. In fact, the relation $\langle a,b;(ab)^2\rangle\cong \mathbb{Z}*\mathbb{Z}_2$ together with the Classification of compact 2-manifolds Theorem, shows that no compact 2-manifold has a fundamental group with $\langle a,b;(ab)^2\rangle$ as a presentation.

As a side note, the space $X$ obtained from a square by identifying the edges according the word $abab$, is homeomorphic to the projetive plane and we can check this by cutting and pasting as is described in the comments. Therefore the fundamental group of $X$ is isomorphic to $\mathbb{Z}_2$. Alternatively, if we use the Seifert-van Kampen Theorem directly in the space $X$, we can choose open subspaces $U$ and $V$ of $X$ such that $X=U\cup V$, $U\cap V$ is homotopicaly equivalent to $S^1$, $U$ is contractible and $V$ is homotopically equivalent to a circunference identifying antipodal points ("the border of the square $X$"), which is isomorphic to $S^1$. Then we will obtain $\langle (ab),(ab)^2\rangle$ as a presentation of $\pi_1(X)$, which is isomorphic to $\mathbb{Z}_2$.

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