Fundamantal group of a regular covering space

Question

Let $B$ be the space of figure $\infty$ (with $x$(the red circle) and $y$(the black one) as generators) and $E$ its covering space (in the picture below).

let $P_{*}: \Pi(E,a) \to \Pi(B,b) $ be the covering map which transformes $a_1 \to x$,$a_2 \to xyx^{-1}$,$a_3 \to x^2yx^{-2}$,$a_4 \to x^3yx^{-3}$,$a_5\to y^4y^{-4}=1$ so $P_{*}( \Pi(E,a))$ seems to be $F_4$. I am locking at the spanning tree of the graph(marked blue in the image) and build the close paths according to their hats which means going on the spanning tree until i am on my generator and coming back so $ \hat{\gamma}_1=a_1 $ and $ \hat{\gamma}_1=\psi * a_2 * \overline{\psi}$ and so on ,so the last one is going on the spanning tree do $a_5$ and coming back from there i will get $1$

But if we calculate this graph fundamental group we will get $F_{8-4+1}=F_5$

Now the covering space is regular so $P_{*}( \Pi(E,a)$ should be normal $F_4$ is indeed normal in $F_2$ but $F_5$ isn't.

I understand that we should calculate $P_{*}( \Pi(E,a))$ according to the Base space $B$ which will give us another relation. But how can it be that $P_{*}( \Pi(E,a)$ is not isomorphic to $\Pi(E,a)$?

Can you explain why does the groups are different? how should i get $P_{*}( \Pi(E,a)$? and if they are how can it be that$P_{*}( \Pi(E,a)$ is not isomorphic to $\Pi(E,a)$?

Answer 1

The homomorphism $p_*$ induced by the covering space map is injective by the unique lifting property. So $\pi_1(E, a)$ is isomorphic to $p_*\left(\pi_1(E, a)\right)$.

The image of $a_5$ (the black circle) is $y^4$. Why do you think it's $y^4 y^{-4}$? The image of a nontrivial loop cannot be trivial by above.

The image of $a_2$ is $yxy^{-1}$. To see this, just map the loop to its image in $B$ and write the image in terms of $x$, $y$.

The subgroup $\langle x, yxy^{-1}, y^2xy^{-2}, y^3xy^{-3}, y^4\rangle$ is indeed normal in $\langle x, y\rangle$. This can be verified by direct inspection. Take the conjugate of each generator by $x$, $y$ and verify that the result is also an element of the same subgroup.