Covering of discrete probability measures

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? I have not been able to disprove it so I'm thinking maybe it is...?

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 $$Y$$."

• So if we replace $\mathcal{P}_{n:+}$ with the set $\left\{ \sum_{i=1}^n k_i\delta_{x_i}:k_i>0,\, x_i\neq x_j \mbox{ if } i\neq j\right\}$ then this should work? – James_T Sep 18 '20 at 7:34
• @James_T : I think so. – Iosif Pinelis Sep 18 '20 at 12:13