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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeMay 18th 2011

at surjective geometric morphism I have spelled out in detail most of the proof of the various equivalent characterizations, and all of the proof of the statement that geometric surjections are comonadic.

1. Added to surjective geometric morphism and to connected topos that any connected geometric morphism is surjective.

2. I added a necessary assumption in the proposition (this assumption is also in MacLane–Moerdijk).

Jens Hemelaer

3. Thanks! I added a reference to the page in MacLaneMoerdijk, a remark on the parallel case of injective maps and specified the precise dependencies in the equivalence.

By the way, does anybody know off hand what the epimorphisms in $\mathbf{Top}_{T_1}$ are ?

Anonymous

• CommentRowNumber5.
• CommentAuthorMike Shulman
• CommentTimeAug 5th 2019

I find the hypothesis “$Y$ is $T_1$” curious. I would have expected instead something like “$X$ is sober and $Y$ is $T_0$”, with the proof proceeding by regarding a point of $Y$ as a point of $Sh(Y)$, lifting it somehow to a point of $Sh(X)$, and using sobriety to extract a point of $X$, whose image in $Y$ is isomorphic to the original one, hence equal by $T_0$. Actually it seems that would only use the “other half” of sobriety for $X$ (existence of points rather than their uniqueness). Is there a chance of such an argument also working?

• CommentRowNumber6.
• CommentAuthorTodd_Trimble
• CommentTimeAug 6th 2019

For what it’s worth, I will answer the question of #4: I will argue that epimorphisms in $Top_{T_1}$ are just the surjections.

Suppose $f: X \to Y$ in $Top_{T_1}$ is not surjective, and take a point $y \in Y \setminus im(f)$. I will argue that every point of the ordinary topological pushout $Y \sqcup_X Y$ is closed, i.e., this pushout object lives in $T_1$ so that the evident fork

$X \stackrel{f}{\to} Y \stackrel{\overset{i_1}{\to}}{\underset{i_2}{\to}} Y \sqcup_X Y$

lives in $Top_{T_1}$. In that case, since $i_1 \circ f = i_1 \circ f$ but $i_1(y) \neq i_2(y)$, we would conclude $f$ is not epic in $Top_{T_1}$.

So we must verify that each point of $Y \sqcup_X Y$ is closed. The points of $Y \sqcup_X Y$ are certain equivalence classes of points of $Y \sqcup Y$, so we verify that the equivalence classes are closed in $Y \sqcup Y$. But each equivalence class consists of either a single point (the case where the point doesn’t “come from $X$”), or two points (a point of the form $f(x)$ living in the first summand $Y$, and the same but living in the second summand). But finite subsets of $Y \sqcup Y$ are closed since this space is $T_1$, so all these equivalence classes are closed, and we are done.

• CommentRowNumber7.
• CommentAuthorThomas Holder
• CommentTimeAug 6th 2019

Thanks, Todd! When I asked I was thinking of surjective geometrical morphisms as “epimorphisms” in $\mathfrak{Top}$ but meanwhile I stumbled over ex.4.4, p.133 in the 1977 elephant stating that they lack the epic cancellation property.