3
$\begingroup$

A map $f : X \rightarrow Y$ is said to be a local homeomorphism if each point $x \in X$ has a neighborhood $U_x$ such that the restriction $f|_{U_x} : U_x \rightarrow f(U_x)$ is a homeomorphism. Let $X$ and $Y$ be path-connected compact Hausdorff spaces and let $f : X \rightarrow Y$ be a surjective local homeomorphism.

(a) Show that any point $y \in Y$ , the set $f^{-1}(\{y\})$ is finite.

(b) Show that for any two points $y_1, y_2 \in Y$ , the sets $f^{-1}(\{y_1\})$ and $f^{-1}(\{y_2\})$ have the same cardinality.

There are many informations and I have no idea how to start with the problem. Here are somethings that I observe:

Each $\#f^{-1}|_{U_x}(\{y\})=1$.

Both $U_x$ and $f(U_x)$ are Hausdorff and path-connected. But They are not necessarily compact.

Please helps.

$\endgroup$
1
$\begingroup$

For $a)$, use the fact the each element in $f^{-1}(y)$ is an isolated point and the fact that $X$ is compact.

For $b)$, take $y \in Y$ and write $f^{-1}(y) = \{x_1,\dots,x_k\}$. Take disjoint open neighbourhoods $U_i$ about each $x_i$ than all map homeomorphically onto the same open neighbourhood $V'$ about $y$ (how?). Let $M = X\backslash(\cup_{i=1}^kU_i)$ and let $N = f(M)$. Let $V = V'\cap(Y\backslash N)$. Then $V$ is an open set, and every point in $V$ has exactly $k$ preimages in $X$ (why?). Thus the map sending points of $Y$ to the number of preimages they have in $X$ is integer-valued and locally constant, so since $Y$ is connected, it must be constant.

Some relevant facts:

Any closed subset of a compact space is compact.

Any compact subset of a Hausdorff space is closed.

The continuous image of a compact set is compact.

Note also that you don't need to assume that $f$ is surjective, any local homeomorphism is an open map so $f(X)$ is open in $Y$. Since $X$ is compact, $f(X)$ is also compact and then $Y$ Hausdorff implies that $f(X)$ is closed. Since $Y$ is connected, it must be that $f(X) = Y$.

I have left out a lot of the details because this is a very important exercise in many areas of maths, including algebraic topology and differential geometry. The example you have given can be generalised or changed in many ways, and very similar results will be true. A similar argument to the above, for example, shows why the notion of valency for an analytic function between Riemann surfaces is well defined. If we suppose that the covering is such that we may lift paths in a unique way (connectedness implies that lifts are unique, but we need more, such as the covering being regular to assume that lifts always exist. If this happens we don't even need the covering space to be connected for uniqueness), then showing that the fibres (sets of pre-images) are all in bijection is a lot easier, and a lot nicer as well. If you know what a regular covering is, you might like to think about it. In fact, I have just realised that the proof outlined above shows that the map $f$ is actually a regular covering, which is a stronger result that you may find useful.

If you have any particular questions about steps you don't understand, leave a comment and I'll be more than happy to try and explain something in greater detail.

$\endgroup$
  • $\begingroup$ Thank you for this. Regarding your first line, what can I read to prove that the inverse image of $y$ consists only of isolated points? Why it cannot have, let's say, limit points? $\endgroup$ – evaristegd May 11 at 3:09
1
$\begingroup$

proof (a) Let $x\in X$ be any point and consider its preimage $f^{-1} (x)$ . This set is compact because $f$ is proper. And since $f$ is a local homeomorphism, it follows the set $f^ {-1}(x)$ is also discrete. Indeed, given any $y\in f^{−1} (x)$ , there is a neighborhood $y\in V$ , such that $f|V :V\to f(V)$ is a homeomorphism and in particular, $y$ is the only preimage of $x$ inside $V$ . Combining these facts, we get that the preimage of $x$ is finite, say, $f^{−1} (x)=\{y_1,\cdots ,y_n \}$.

$\endgroup$
  • $\begingroup$ Why do you conclude that $y$ is the only preimage of $x$ inside $V$? $\endgroup$ – evaristegd May 11 at 3:20
0
$\begingroup$

A map $f$ with such property is a covering, hence bot results follow directly from well-known facts from algebraic topology (e.g. from the theory of covering of groupoids).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.