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In page 70 of Hatcher's book, in the section Deck Transformations and Group Actions, the author defines a normal covering in the following way:

A covering space $p:\tilde X\to X$ is called normal if for each $x\in X$ and each pair of lifts $x_1,x_2$ of $x$ there is a deck transformation taking $x_1$ to $x_2$.

Afterwards, the author says this covering of $S^1 \vee S^1$

enter image description here

is a normal covering, I know why this is a covering, but I didn't understand why this is a normal covering based in this definition.

Please I really need help

Thanks a lot.

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The flipping of the b edges and rotating the diagram by 180 degrees are both deck transformations. –  Sanchez Jan 13 '13 at 1:35
    
@Sanchez What do you mean by "the flipping of the b edges"? thank you for your comment. –  user42912 Jan 13 '13 at 1:42
    
Sorry, that part was nonsense. Rotation should be sufficient. –  Sanchez Jan 13 '13 at 1:45
    
@Sanchez Sorry, but I still don't understand why this covering in normal according to the Hatcher's definition :( –  user42912 Jan 13 '13 at 2:07
    
For a generic point of $S^1 v S^1$, what are the lifts? –  Sanchez Jan 13 '13 at 2:10
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1 Answer

By proposition 1.39(a) of Hatcher it will suffice to show that

$$H = \pi_1(\tilde{X}) = \langle a,b^2, bab^{-1} \rangle \unlhd \langle a,b\rangle.$$

$\tilde{X}$ is the covering space in your picture above. Now to show normality it will suffice to conjugate each of the generators of $H$ by $a$ and $b$ and see if they're in $H$. Clearly any conjugate of $a$ is in $H$. Any conjugate of $b$ is also in $H$ because $b^{-1}b^2b = b^2$ while $ab^2a^{-1}$ has to be in $H$ because $a$ and $a^{-1}$ are. If you conjugate $bab^{-1}$ by $b$ you get $b^2a(b^{-1})^2$ which is in $H$ because $b^2,a,b^{-2} \in H$ while if you conjugate by $a$ you also get something in $H$ because $a$ is already in $H$.

Hence the covering space $\tilde{X}$ in your picture above is a normal covering space of $S^1 \vee S^1$.

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