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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...?

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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 $Y$."

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  • $\begingroup$ 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? $\endgroup$
    – ABIM
    Commented Sep 18, 2020 at 7:34
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    $\begingroup$ @James_T : I think so. $\endgroup$
    – Iosif Pinelis
    Commented Sep 18, 2020 at 12:13

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