By / 18 July, 2022

Q: Does the image of a weakly positive $V$-filtered object under a $V$-filtering monad form a weakly positive $V$-filtered object? It is not difficult to check that the image of a weakly positive $V$-filtered object under a $V$-filtering monad $T$ is again weakly positive. Question: Given $T$ a $V$-filtering monad, if $X$ is a weakly positive $V$-filtered object, is the image $TX$ a weakly positive $V$-filtered object? A: One way of seeing that it is false in general is that if $A$ is a strongly positive $V$-filtered object and $X$ is a weakly positive $V$-filtered object then $TX$ is just $A$ with a new filtering sieve, but $A$ is not weakly positive. This is because the fiber over $a\in A$ is not a product of positive $V$-filtered spaces when $A$ is not a product of positive $V$-filtered spaces. Another way of looking at it is by using the theorem that a closed monomorphism between weakly positive $V$-filtered objects is a filtered monomorphism. But $TX$ is not closed under forming filtered colimits because it is not a product. Also, even when the $V$-filtering property is unnecessary, the normal condition on $\varphi:TX\to B$ as a functor between filtered spaces which is a closed filtered monomorphism is equivalent to the statement that for each $a\in A$, the natural map $\varphi_a:TX\to B$ preserves filtered colimits in the $TX$ variable. But in general this is not true: let $X$ be the filtered space on a singleton and $A\to X$ be the inclusion. Then the natural map $TX\to B$ is just the fiber map $A\to B$ over $x\in X$. But $TX$ is the colimit of $A\to X\to B$ while $A$ is not filtered over any sieve. On the other hand, if $A$ is strongly positive, then $TX$ has the
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