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I am studying some properties of local homeomorphism

I am in particular trying to find a local homeomorphism that is not a homeomorphism and the projection function seems to be the perfect candidate since it doesn't "recover".

I found an example here:

For $X$ any topological space and for $S$ any set regarded as a discrete space, the projection

$\pi_X: X\times S \to X$ is a local homeomorphism.

I am trying to see how this function fits the definition:

Definition: Let $(X, \mathcal{T})$ and $(Y, \mathcal{J})$ be topological spaces. A function ${\displaystyle f:X\to Y\,}$ is a local homeomorphism if for every point $x \in X$ there exists an open set $U \subseteq X$ containing $x$ and an open set $V \subseteq Y$ such that the restriction ${\displaystyle f|_{U}:U\to V\,}$ is a homeomorphism.

So let $V_x$ be an open set on $X$, and $U_x$ some open set in $X \times S$ containing $x$, we wish to show that $\pi_X|_{U_x}: U_x \to V_x $ is a homeomorphism

  1. Show $\pi_X|_{U_x}$ is continuous: Take an open set in $V_x$, then such set has the form $V_x \cap W$, $W$ is open in $X$, $\pi_X|_{U_x}^{-1}(V_x \cap W) = \pi_X|_{U_x}^{-1}(V_x) \cap \pi_X|_{U_x}^{-1}(W) = \pi_X|^{-1}(V_x) \cap U_x \cap \pi_X|^{-1}(W)$, right handside is open in the product topology on $X \times S$

  2. Show $\pi_X|_{U_x}$ is open: Take an open set in $U_x$, then such set has the form $U_x \cap M$, $M$ is open in $X \times S$, $\pi_X|_{U_x}(U_x \cap M) = \pi_X|_{U_x}(U_x) \cap \pi_X|_{U_x}(M) = \pi_X( U_x) \cap\pi_X(M) \cap \pi_X( U_x)$, right handside is open since $\pi_X$ is open map$

[^ actually I noticed that we cannot do $f(A \cap B) = f(A) \cap f(B)$ since $\pi$ is not a bijection...]

Further, this map is not a homeomorphism since $\pi_X$ does not biject.

Is the above good enough? Is there any easier way to show this result?

Note: another example that seems more accessible is $f: \mathbb{C} \to \mathbb{C}$, $f(z) = e^z$, but I don't know about topologies on complex spaces

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  • $\begingroup$ If $X$ is empty or if $S$ has exactly one member then $\pi_X:X\times S\to X$ is a homeomorphism $\endgroup$ – DanielWainfleet Jun 15 '16 at 20:27
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You have given an argument that $\pi_X|_{U_x}$ is continuous and open, but you are missing the most important part: a homeomorphism needs to be a bijection! In fact, $\pi_X|_{U_x}$ won't be a bijection if you choose $U_x$ and $V_x$ arbitrarily (for instance, you might choose $U_x$ to be all of $X\times S$ and $V_x$ to be all of $X$, and then $\pi_x$ is not a bijection unless $S$ has exactly one point.)

So you need to use the fact that the definition of "local homeomorphism" lets you choose $U$ and $V$. Following the definition, you should not start with $V_x$ and $U_x$, but only with a point $(x,s)\in X\times S$. You then get to choose an open set $U$ containing $(x,s)$ and an open set $V$ such that $\pi_X$ will restrict to a homeomorphism $U\to V$. Let's choose $V=X$ and $U=X\times \{s\}$. Note that $U$ is open because $S$ has the discrete topology so $\{s\}$ is open in $S$. Then it is easy to show that $\pi_X$ restricts to a bijection $U\to V$, and your argument shows this bijection is a homeomorphism.

In fact, if you just want any example at all, there is a super-simple one: just consider the map $\emptyset\to Y$ for any nonempty space $Y$. This is trivially a local homeomorphism, since there are no points in the domain to choose when checking the definition!

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  • $\begingroup$ So you mean consider the space $\{\varnothing\}$ and $Y$, $Y$ nonempty, then the map $\varnothing \to Y$ is a local homeomorphism. $\endgroup$ – Carlos - the Mongoose - Danger Jun 15 '16 at 18:33
  • $\begingroup$ The space is $\emptyset$, not $\{\emptyset\}$, but yes. $\endgroup$ – Eric Wofsey Jun 15 '16 at 18:36
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The sets $X_s = X \times \{s\} \subset X \times S$ are all open since $S$ is discrete, and $\pi_X|_{X_s} \colon X_s \to X$ is a homeomorphism (the inverse is $x \mapsto (x,s)$). Since $\{X_s\}_{s\in S}$ form an open cover of $X\times S$, we have shown $\pi_X$ is a local homeomorphism.

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