Let $\mathcal{P}\_{n:+}(\mathbb{R})$ denote the set of probability measures on $\mathbb{R}$ for the form $\sum\_{i=1}^n k\_i \delta\_{x\_i}$ where $k\_i>0$. Then any measure in $\mathcal{P}\_{n:+}(\mathbb{R})$ is in the image of the map on $\Delta\_n \times \mathbb{R}^n$, where $\Delta\_n$ is interior of the $n$-simplex (i.e.: $k\_1,\dots,k\_n \in (0,1)$ with $\sum\_{i=1}^n k\_i =1$, taking $(k\_1,\dots,k\_n)\times (x\_1,\dots,x\_n)$ to $\sum\_{i=1}^n k\_j \delta\_{x\_i}$. Clearly this map is continuous, when $\mathcal{P}\_{n:+}(\mathbb{R})$ is equipped with the Prokhorov metric.


However, is it a *[covering map](https://mathworld.wolfram.com/CoveringMap.html)*? I have not been able to disprove it so I'm thinking maybe it is...?



Covering of discrete probability measures
This map is not a covering one, because the preimages of singleton sets under this map are not of the same cardinality. E.g., the cardinality of the preimage of the singleton set $\{\frac1n\,\sum\_{j=1}^n\delta\_j\}$ is $n!$, whereas the preimage of the singleton set $\{\delta\_0\}$ is of infinite cardinality.




---


From the site you linked to: "[for a covering map $f\colon X\to Y$,] the cardinal number of $f^{-1}(y)$ (which is possibly infinite) is independent of the choice of $y$ in $Y$."


