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comment algebraic closedness in in residue field
I was going to answer this at MSE but unfortunately the migration got rejected before I could. Any $x/y\in Q(C)$ is defined on a dense open subset of almost every fiber of $\operatorname{Spec} A\times \operatorname{Spec} B\to \operatorname{Spec} A$. If $x/y$ is algebraic over $A$, then its restriction to almost every fiber will be algebraic over $k$ and hence constant.
Apr
21
comment Categories in which an epimorphism applied to a non-monic epimorphism can be monic
What sort of thing do you hope to be able to conclude about $\mathcal{C}$? Note that it is possible for $\mathcal{C}$ to be complete and cocomplete: for instance, in $\mathcal{C}=CRing^{op}$, take $A=k(x)$, $B=k[x]$, and $C=k[x^2,x^3]$ with the obvious maps.
Apr
19
revised Minimal totally separated spaces
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Apr
19
comment Minimal totally separated spaces
@DominicvanderZypen: It's not true in general, but I don't see how that's relevant to my answer. I'll go ahead and add in the proof that minimal totally separated is equivalent to Stone to avoid any confusion.
Apr
19
comment (co)limits in the category of diffeological spaces vs. category of smooth manifolds
Manifolds certainly do not form a reflective subcategory, but some individual diffeological spaces still have reflections (i.e., an initial object in the category of manifolds under them).
Apr
18
comment Minimal totally separated spaces
The original version of this answer was wrong because it did not guarantee that $B$ was actually the entire clopen algebra of $X$; I have corrected it (at the cost of some generality).
Apr
18
revised Minimal totally separated spaces
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Apr
18
revised Minimal totally separated spaces
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Apr
18
comment Minimal totally separated spaces
@DominicvanderZypen: An easy way to get a non-$T_2$ totally disconnected space is to take a totally disconnected space and double a non-isolated point.
Apr
18
comment Minimal totally separated spaces
@AndrejBauer: I don't understand your last comment. (Also, thanks for the edit :) ).
Apr
18
comment Minimal totally separated spaces
I just meant that for the purposes of the question, you might as well assume clopens form a basis, since if they don't then the topology they generate is a smaller 0d topology.
Apr
18
answered Minimal totally separated spaces
Apr
18
comment Minimal totally separated spaces
@DominicvanderZypen: Every 0d space naturally injects into the Stone space of its clopen algebra, and if this map is not surjective then my first comment describes how to find a subalgebra that still separates points.
Apr
18
comment Minimal totally separated spaces
A m0d space is the same thing as a Stone space (if some ultrafilter $U$ on the clopen algebra is not represented by a point of $X$, then the subalgebra that identifies $U$ with some point of $X$ still separates points of $X$).
Apr
18
comment (co)limits in the category of diffeological spaces vs. category of smooth manifolds
A counterexample to the conjecture in my previous comment: let $M$ be any manifold and let $X$ be $M$ equipped with the diffeology generated by smooth paths. Then the identity $X\to M$ realizes $M$ as the manifold reflection of $X$ (by Boman's theorem), is a homeomorphism, and is not a diffeomorphism unless $\dim M\leq 1$.
Apr
18
comment (co)limits in the category of diffeological spaces vs. category of smooth manifolds
As Martin Brandenburg mentioned on MSE, non-Hausdorff manifolds give examples of colimits that are not preserved. It seems plausible to conjecture that any Hausdorff diffeological space that has a manifold reflection is already a manifold.
Apr
18
comment (co)limits in the category of diffeological spaces vs. category of smooth manifolds
All limits are trivially preserved, since a diffeological space is determined by the maps from manifolds into it. Since every diffeological space is the colimit of all the manifolds mapping into it, there is a colimit that is not preserved iff there is a diffeological space which is not a manifold but has a reflection in the subcategory of manifolds.
Apr
17
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Apr
17
comment Connected, maximal compact, but not $T_2$
The one-point compactification will not be maximal compact unless $Y$ is compactly generated. Indeed, if $A\subset Y$ has closed intersection with every compact set but is not closed, then $A\cap \{\infty\}$ is compact but not closed in the one-point compactification.
Apr
17
answered Connected, maximal compact, but not $T_2$