I've read (and I see it more or less clearly) that a 2-folded covering space $\tilde{X}$ of $X=S^1 \vee \mathbb{R}P^2$ is a 2-sphere with 2 unit circles tangently attached to antipodal points. If my reasoning is correct, $\pi_1(\tilde{X}) = \langle a, b \rangle$ is generated by two loops, each one around each circle, while $\pi_1(X) = \langle c, d \mid d^2=1 \rangle$ ($c$ loops around the circle, $d$ is the generator of the group of the plane).
Now, the covering map $\pi$ between them induces an injective homomorphism between the fundamental groups. But in this case, the two circles map to the same one in $X$, so $a$ and $b$ both map to $c$. Where is the error, in the covering spaces themselves or my calculation of their fundamental groups? Any help is appreciated.
Thank you in advance :)
EDIT: I include a picture showing how I think the covering would work. Points in $S^2$ get mapped to their projection on the plane, and points in both $S^1$ go to the same position in the $S^1$ of the image.
You seem to have switched around $X$ and $\overline{X}$: $\pi_1(X)\cong\langle c,d|d^2=1\rangle$, rather, and $\pi_1(\overline{X})\cong\langle a,b\rangle$.
So the trouble must be that the map $p:\overline{X}\to X$ which is given by the components $S^1=S^1$ and $S^2\twoheadrightarrow\Bbb RP^2$ is not actually a covering map. Consider, say, any neighbourhood of the base point of $S^1$ (which is forced to be shared by both copies of $S^1$). It pulls back under $p$ to two identical but nondisjoint copies in $\overline{X}$ and it is completely impossible to have the "disjoint sheets" condition met. So, $p$ is not a covering.
EDIT: I now see you have defined your space $\overline{X}$ differently to how I expected from the notation $S^1\vee S^2\vee S^1$. The map is now a covering but it does not send both $a,b$ to $c$. In this image:
Observe that while $a\mapsto c$ is true, the loop class associated to $b$ - which must be about the same base point as $a$ - actually crosses into a loop around $\Bbb RP^2$ after projection by $p$. Note the purple line around $S^2$ which runs $\beta\to\alpha$: this projects to a generator of $\pi_1(\Bbb RP^2)$. So, $b\mapsto dcd$. And the homomorphism induced by $(a,b)\mapsto(c,dcd)$ is an injection. So all is well!