Computing deck transformation groups I have a problem with computing the deck transformation groups.

Let $\pi: Y \to X$ be a covering map.
a) Show that the map
$f: C \to C\setminus \{0\} , f(z)=e^z\tag{1}$
and all quotient maps
$g_n: S^n \to RP^n\tag{2}$
are covering maps. Which of them are universal?
b) Assume now that $\pi: (Y,y_0) \to (X,x_0)$ is a   universal covering with $\pi(y_0)=x_0$. Let $Aut_X(Y)$ denote the homeomorphisms that commute with $\pi$, i.e.  $Aut_X(Y):=\{f: Y \to Y : \pi \circ f=\pi, f \ is \ a \ homeomorphism \}$.
Deduce the universal covers and compute $Aut_X(Y)$ for $X=C\setminus \{0\}$ and $X=RP^n$ for $n \geq 1$, c.f. part a).
Hints: You may use (without proof) that an $f \in Aut_X(Y)$ is uniquely given by $f(y_0)$ and that $Aut_X(Y)$ comprises the same information as $\pi_1(X)$, throughout given that X is sufficiently nice (connected and locally simply connected) which you may assume as well.

I have already solved a) and found out that the map (1) is a universal covering and that map (2) is universal if $n \geq 2$.
For b) consider the case $X=C \setminus \{0\}$. If $Y=C$ I can find a universal cover. Any homeomorphism $f \in Aut_X(Y)$ must satisfy $e^{f(z)}=e^z$. So $f(z)=e^{i2 \pi m z}, \ m \in Z$, where $Z$ denotes the group of integers. This implies that $Aut_X(Y)$ is isomorphic to $Z$.
Now consider the case $X=RP^n$. If $Y=S^1$ we can distinguish to cases. If $n=1$ then $RP^n$ is homeomorphic to $S^1$ so $\pi_1(RP^n)$ is isomorphic to $\pi_1(S^1)$ and since $\pi_1(S^1)$ is isomorphic to $Z$ it follows from the hint that $Aut_X(Y)$ is isomorphic to $Z$. If $n \geq 2$ we know that any homeomorphism $f \in Aut_X(Y)$ must satisfy $[f(z)]=[z]$ where $[z]$ denotes an equivalence class contained in $RP^n$. Since $RP^n$ is obtained from $S^n$ by identifying antipodal points and $f(z)$ and $z$ are equivalent we have $f(z)=-z$ or that $f$ is the identity map. This implies that $Aut_X(Y)$ is isomorphic to $Z_2$.
However I am not sure whether this is sufficient to compute the deck groups. I also don't know how to use the hint to deduce the universal covers since I only know the definition of covering maps and nothing else.
 A: Your approach is correct. Here are some comments.


*

*$e^{f(z} = e^z$ means $e^{f(z) - z} = 1$, i.e, $f(z) - z = 2m\pi i$. Thus $f(z) = z + 2m\pi i$. That is, $f$ is a translation of the complex plane by an integral multiple of $2\pi i$.

*In fact $RP^1 \approx S^1$. You have to invoke that $p : R \to S^1, p(t) = e^{i t}$, is the standard  universal cover of $S^1$. The hint does not give you this information. Then it is easy to see that $f$ must have the form $f(t) = t + 2m\pi$.

*In the above two cases this automatically gives you the group structure of $Aut_X(Y)$. In fact, $Aut_X(Y) \approx Z$.

*The only place where you need the hint is $X = RP^n$ for $n > 1$. You have shown that $Aut_X(Y)$ has exactly two elements: The identity and the antipodal map. Thus $Aut_X(Y) \approx Z_2$. However, you have a little gap: You know that  $[f(z)] = [z]$, but this only shows that $f(z) = \pm z$. You can either use a continuity argument to conclude that $f(z) = z$ or $f(z) = -z$ for all $z$ or use the hint ($f$ is uniquely determined by a single $f(y_0)$).

*It seems that you have doubts whether the hint suffices to detect the group structure of $Aut_X(Y)$. It depends on how you interpret the hint. I believe it is too vague, especially the phrase "$Aut_X(Y)$ comprises the same information as $\pi_1(X)$". In fact it is true that $Aut_X(Y) \approx \pi_1(X)$ for sufficiently nice $X$, but if you know this, then you do not need to determine explicitly what the universal cover looks like.
