Equivalent presentation for the fundamental group of the projective plane

Question

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?

Answer 1

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 $\mathbb{Z}_2$.