Can a continuous topological function be one-many?

Question

Can there exist a continuous topological mapping $f:X\to Y$ which is one-many?

I ask this question because if such a mapping exists, then I can see potential contradictions in some theorems stated in my text.

I know that one-many mappings aren't called functions in real analysis. I want to know if the same principle continues even in topology. I searched a lot online, but could not find any such statement.

Thanks in advance!

I think dismissing the question because "functions" are single-valued misses the point of the question. — Austin Mohr, Jun 03 '13 at 07:48
Related: http://mathoverflow.net/questions/179123/continuous-relations . — Michał Kukieła, Aug 23 '14 at 10:13
possible duplicate of Can we extend the definition of a continuous function to binary relations? — Bartek, Aug 28 '14 at 10:43

score 7 · Answer 1 · answered Jun 03 '13 at 07:31

To accompain the given answers, there is a theory of so called multivalued mappings $f\colon X \multimap Y$ being defined as maps $f \colon X \to \mathcal P(Y)$ and continuity for such multis, which generalizes the known definition for single valued maps, where the latter are interpreted as "multis" taking values in $\bigl\{\{y\} \mid y \in Y\bigr\}$.

For a multi $f \colon X \multimap Y$ and $A \subseteq Y$ we define $$ f_+^{-1}(A) = \{x \in X \mid f(x) \ne \emptyset, f(x) \subseteq A\}$$ and $$ f_-^{-1}(A) = \{x \in X \mid f(x) \cap A \ne \emptyset \}$$ A multi is called upper resp. lower semicontinuous if $f_+^{-1}(A)$ resp. $f_-^{-1}(A)$ is open in $X$ for each open $A \subseteq Y$. $f$ is called continuous if it is both upper and lower semicontinuos. Some theorems for continuous maps $f \colon X \to Y$ have their generalisation for continuous multis $f \colon X \multimap Y$.

This seems to match the definitions in Claude Berge's Topological Spaces (which I was about to edit in), so instead I will just delete my answer. — user642796, Jun 03 '13 at 07:36

score 2 · Answer 2 · answered Jun 03 '13 at 07:02

A function from $X$ to $Y$ by definition takes each $x\in X$ to a single $f(x)\in Y$. One can of course have a function $f:X\to\wp(Y)$, in which case each $f(x)$ is a subset of $Y$, but this is not a function from $X$ to $Y$: one would not write $f:X\to Y$. This has nothing to do with real analysis or topology: it’s part of the definition of function.

score 2 · Answer 3 · answered Jun 03 '13 at 07:07

This definition of function is from general topology by Engelking:

Any subset of the Cartesian product $X \times Y$ is a relation. The relation $f\subset X\times Y$ is called a function from $X$ to $Y$, or a mapping of $X$ to $Y$, if for every $x\in X$ there exists a $y \in Y$ such that $(x,y)\in f$ and $y$ is uniquely determined by $x$, i.e., $(x,y) \in f$ and $(x,y') \in f$ imply $y=y'$.

So I am sorry to say the answer is No.

Lehs · Answer 4 · 2014-08-23T11:01:59.070

If $R$ is a binary relation (~unproper function) $R\subset X\times Y$ there is a definition of continuity that extends the definition for (proper) functions:

$R\subset X\times Y$ is continuous iff for all sets $M\subset Y$ it holds that $ xRy\wedge$ $x\in \overline{R^{-1}(M)} \Rightarrow y\in \overline{M}$

It is very intuitive and corresponds to the original $\delta$-$\varepsilon$ technique in calculus. This formula was the result of a construction of a category of mathamatical structures, so I haven't invented it, and I can't guarantee that it is relevant. You can probably not prove any contradictions with it, because it is only a definition, but you might be able to pinpoint to relevant distinctions (or something like that) by using it.

Can a continuous topological function be one-many?

4 Answers4