It is known that given a uniform space $(X, \mathcal{U})$ and a function $\pi$ from $X$ onto an arbitrary set $Y$, it is possible to define a uniformity on $Y$, in the following way: \begin{equation*} \mathcal{U}_\pi= \{B: B\subseteq Y\times Y: \pi^{-1}(B)\in\mathcal{U}\}. \end{equation*} The uniformity $\mathcal{U}_{\pi}$ is called the quotient uniform with respect to $\pi$ and also we can say that \begin{equation*} \mathcal{U}_{\pi}=\{\pi\times \pi(U): U\in\mathcal{U}\}. \end{equation*}

Q. Is it true that for every $U\in\mathcal{U}$, there is $V\in\mathcal{U}_\pi$ such that if $(\pi(x),\pi(y))\in V$, then $(x, z)\in U$ for some $z\in\pi(y)$.

  • $\begingroup$ Are you sure $\mathcal{U}_\pi$ is a uniformity ? What if $f$ is constant equal to $y$ and $B=\{(y,y)\}$ ? $\mathcal{U}_\pi$ will be a filter on $Y\times Y$ but it will not satisfy the conditions that make it a uniformity in general $\endgroup$ – Max May 7 '18 at 6:14
  • $\begingroup$ @max, $f$ is onto. $\endgroup$ – user479859 May 7 '18 at 11:56
  • $\begingroup$ You need some extra condition on $\pi$ to be able to define a "co-induced" uniformity. $\endgroup$ – Henno Brandsma May 9 '18 at 22:52

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