I am trying to solve the following exercise by two different ways:
Compute the fundamental group $\pi_1(\mathbb R\mathbb P^2-\{p\})$
For do so, first we need to remember that $\mathbb R\mathbb P^2$ can be constructed from the closed 2-disk $D^2\subset \mathbb C$ together with the relation equivalence generated by $z\sim -z$ on its boundary $\partial D^2\cong S^1$.
By Seifert-Van Kampen's theorem. Let $A$ and $B$ be the open subsets of $\mathbb R\mathbb P^2$ defined as the image of the quotient map $D^2\to\mathbb R\mathbb P^2$ of the subsets in $D^2$ given by $|z|<3/2$ and $|z|>1/2$ respectively. It can be proved that $A$, $B$ and $A\cap B$ are indeed open path-connected subset of $\mathbb R \mathbb P^2$ and are homotopic equivalent to $S^1$. Each one-degree closed loop in $A\cap B$ wraps $A$ once and wraps $B$ two times, then Seifert-Van Kampen's Theorem (p. 43 of the Hatcher's book) ensures: $$\pi_1(\mathbb R\mathbb P^2-\{p\})\cong \frac{\langle a,b\rangle}{\langle\langle a-b^2\rangle\rangle}=\langle a,b|a=b^2\rangle\cong \mathbb Z.$$
By considering deformation retractions. The maps $h_t:D^2-\{p\}\longrightarrow D^2-\{p\}$ given by $h_t(z)=(1-t)z+tz/|z|$ defines a deformation retraction (in sense of p.2 of Hatcher's book) from $D^2-\{p\}$ to $S^1$ that induces an homotopy in the quotient space in a natural manner. This implies that $\mathbb R\mathbb P^2-\{p\}$ has the same homotopy type of $S^1/\sim$, being $\sim$ the equivalence relation described above. It is pretty clear that $S^1/\sim$ is homeomorphic to $S^1$. Then $\pi_1(\mathbb R\mathbb P^2)\cong\mathbb Z$.
Both anwers gives us the same solution $\pi_1(\mathbb R\mathbb P^2-\{p\})\cong \mathbb Z$. I feel comfortable with the first one (using Van-Kampen's Theorem) but in the second one I think I am making a mistake when I define the deformation retraction in the quotient space $\mathbb R\mathbb P^2-\{p\}$.
I will be grateful if someone confirm me that both solutions are correct.