Distinct topologies making a map a local homeomorphism

Question

If $X$ is a topological space and $p:E\rightarrow X$ is a function, is it possible for $E$ to have two distinct topologies each one making $p:E\rightarrow X$ a local homeomorphism?

Obs.: A local homeomorphism $f:A\rightarrow B$ between topological spaces is a function such that for every $a\in A$ there are open sets $U$ and $V$ in $A$ and $B$ such that $a\in U$ and $f\upharpoonright U:U\rightarrow V$ is a homeomorphism.

Answer 1

Yes. Consider $E=\{1,2,3,4\}$ and $X=\{1,2\}$. Let $p:E\to X$ be given by

$$p(1)=p(2)=1$$ $$p(3)=p(4)=2$$

Now put the antidiscrete topology on $X$: $\tau_X=\{\emptyset, X\}$. While on $E$ put these two topologies:

$$\tau_{E,1}=\big\{\emptyset, E, \{1,3\}, \{2,4\}\big\}$$ $$\tau_{E,2}=\big\{\emptyset, E, \{1,4\}, \{2,3\}\big\}$$

Note that $\tau_{E,1}\neq\tau_{E,2}$. But you can easily check that $p$ is a local homeomorphism in both of them.

Side note: these two topologies are distinct (as requested) but homeomorphic. An interesting question is whether we can find non-homeomorphic topologies. I don't know the answer to that.

Answer 2

If $p$ is injective, the two topologies must coincide.

Proof. Composition of local homeo is local homeo. So the identity map, which is the composition with $p$ and its local inverse, is a local homeo between the two topologies. It follows that one is finer than other and viceversa.

With more details, let $\tau,\sigma$ be two topologies on $E$ with requested property. Let $A\in \tau$. Let $x\in A$. By definition there is $U\in \tau$ and $V$ open set in $X$ such that $p:U\to V$ is homeo. So the restriction of $p$ to $A\cap U$ is homeo with its image, which is an open set in $X$ that we name $W_A$.

Similarly for any $B\in \sigma$ containing $x$ we define $W_B$. Now let $W=W_A\cap W_B$ and denote by $P$ the inverse of $p$ restricted to $W$. This is well defined because $p$ is injective.

$W$ is an open set in $X$ containing $p(x)$ and $P(W)$ is a set containing $x$, open in $\sigma$, and contained in $A$.

Therefore $A$ is a $\sigma$-neighborhood of $x$. Since this holds true for any $x\in A$ it follows that $A$ is open in $\sigma$. We have then proved that $A\in\tau\Rightarrow A\in\sigma$. Thus $\tau\subset\sigma$. Interchanging the roles of $\tau$ and $\sigma$ we get the ohter inclusion.