Why do the definition of deck transformations requires homeomorphism This question arose in a first-year graduate topology course. But not a homework!
Recall, given a covering map $p:\tilde{X}\rightarrow X$ of path connected spaces. A deck transformation should be defined as a homeomorphism $f:\tilde{X}\rightarrow\tilde{X}$ satisfying $p(f(x))=p(x)$.
The teacher made a mistake in the class. He only required $f$ to be a continuous map. So I am wondering why continuous is not enough (Since no textbook use it as definition, there should be a counterexample I think). I showed $f$ must be surjective and open, but cannot make it further.
Thank you!
 A: Assume given a group $G$, a subgroup $H$ and a $g\in G$ such that $gHg^{-1}$ is
properly contained in $H$. Let now $Y$ be a topological space with an
action of $G$ such that $Y\rightarrow Y/G=:X$ is a covering map (such $Y$'s are
plentiful). Let $\tilde X:=Y/H$ so that $p\colon\tilde X\rightarrow X$ is
another covering map. Now, $g\cdot\colon Y\rightarrow Y$ maps $H$-orbits to
$H$-orbits and hence induces a map $f\colon\tilde X\rightarrow\tilde X$ which
fulfils $p\circ f=p$.  It is not injective however as the fibres look like
cosets $H/gHg^{-1}$ which by assumption contains more than one element. (Using
some covering space theory it is easy to see that possibly - I don't think this
possibility exists - excluding some very strange examples any example must
appear in this way.)
It remains to show that such $(G,H,g)$ exist. One can actually start with any
$H$ and an injective non-surjective endomorphism of it but it is easier to give
a concrete example (the general construction is very similar). Hence we let
$H':=\mathbb Z[1/2]$, the group of rational numbers whose denominators are
powers of $2$. We have an action of $\mathbb Z$ on $H'$ where $1\in\mathbb Z$
acts by multiplication by $2$. We then let $G$ be the semi-direct product of
$H'$ and $\mathbb Z$, $H:=\mathbb Z\subseteq H'$ and $g=(1,0)$. Note that this group is
finitely generated (by $(1,0)$ and $(0,1)$) but it seems not to be finitely
presented. However, it is easy enough to modify it to be finitely presented:
Just take the group generated by $g$ and $h$ and relation $ghg^{-1}=h^2$ with
$H$ generated by $h$. It maps to the semi-direct product which shows that
$gHg^{-1}$ is indeed properly contained in $H$. Hence we can get examples where
$X$ is a finite CW-complex.
Addendum: I just saw Benjamin's answer when I posted this. He makes use of the claim that any $G$-endomorphism of a transitive $G$-set is an automorphism. However, this claim seems to be false and under my assumptions we get an example of an endomorphism of $G/H$ that is not bijective. Curiously enough my first reaction was the same as Benjamin's. Maybe both of us encountered such statements when learning of finite groups where of course it is true.
Addendum 1: In a comment to Benjamin's answer (that may disappear as Benjamin has  deleted his answer) Georges Elencwajg makes a reference to page 179 of Lima's Fundamental Groups and Covering Spaces where a counterexample to bijectivity is given where $X$ is the figure eight with a $2$-cell adjoined. Let me further add that my last example with the group generated by $g$
and $h$ is most likely to be the same example. We can indeed construct a CW-complex with my example as its fundamental group by
starting with the figure eight and then adding the relation $ghg^{-1}=h^2$ by
adjoining the appropriate 2-cell.
