Let $A\subseteq C$ be the annulus given by $A=\left\{z|1\geq|z|\geq\frac12\right\}$. Define an equivalence relation on $A$ as follows: two different points $z, w$ are equivalent if $|z| = |w| = 1$ and $z = −w$ or if $|z| = |w| =\frac12$ and $z = −w$. Let X = $A/ ∼$. Compute $\pi_1(X)$.
I've tried to use Van-Kampen theorem, by defining $U=\left\{z|\frac34\ >|z|\geq\frac12\right\}$ and $V=\left\{z|1\geq|z|>\frac23\right\}$ Then $\pi_1(U) \cong \mathbb{Z}$ $\cong$ $\pi_1(V)$ and same for the intersection. Then I think that $\pi_1(X)=\mathbb{Z}*\mathbb{Z}/\langle a,b|a^2=b^2\rangle$ Am I correct?
any help will be greatly appriciated
You have the right idea, but as AP says, you have to take the open cover $\{U, V\}$ given by the images of $\{z : 3/4 > |z| \geq 1/2\}$ and $\{z : 1 \geq |z| > 2/3\}$ under the quotient map.
In that case, $U$ and $V$ are both homotopy equivalent to $S^1$, and $U \cap V$ deformation retracts to $S^1$. The inclusion induced maps $\pi_1(U \cap V) \stackrel{\pi_1 i_U}{\longrightarrow} \pi_1(U)$ and $\pi_1(U \cap V) \stackrel{\pi_1 i_V}{\longrightarrow} \pi_1(V)$ multiplies the generator $\times 2$ (geometrically, push the circle in the intersection up towards any of the top or bottom until it wraps twice about itself)
Thus, by applying van Kampen theorem, we get $\pi_1(X) \cong \pi_1(U) * \pi_1(V)/\langle i_U i_V^{-1}\rangle$ which is isomorphic to $\Bbb Z*\Bbb Z/\langle a^2b^{-2}\rangle \cong \langle a,b | a^2 = b^2\rangle$.
On the other hand, a good thing to note will be that identifying the top and bottom circle in the cylinder (which is homeomorphic to an annulus) via $z \sim -z$ is the same as pasting two disks to the two circles via the attaching map $z \mapsto z^2$.
That said, $X$ is a sphere with two cross caps, which, by an appropriate cell decomposition, has $\pi_1 \cong \langle a, b | a^2b^2 = 1\rangle$ which is isomorphic to the group we've wrote down above by the transformation $b \mapsto b^{-1}$.