The Stacks project

Should "tensor product presheaf" be in italics?

Consider the following map of finite topological spaces, hopefully in suggestive notation: $\pi: \{ \bullet_u \rightarrow \star_{a=b} \} \uplus \{ \bullet_{v} \rightarrow \star_{b} \} \longrightarrow \{ \bullet_u \rightarrow \star_{a=b} \leftarrow \bullet_{v} \}$

Here $\uplus$ denotes the disjoint union, the subscripts indicate where each point goes, and $\bullet_u \rightarrow \star_a$ means that point $\star_a$ lies in the closure of point $\bullet_u$.

Then we can add the following item to Proposition 5.26.6(08YN). I state it in full:

Proposition 5.26.6' (08YN'). Let X be a topological space. The following are equivalent (1) X is extremally disconnected, (2) for any solid commutative diagram $X \longrightarrow Z_0 \xleftarrow \pi Y_0$ of continuous maps with the surjective proper map $\pi:Y_0\longrightarrow Z_0$ of finite topological spaces defined above, there is a dotted arrow $X\longrightarrow Y_0$ in the category of topological spaces making the diagram commute.

Proof. It is clear that (1) implies (2). On the other hand, if (2) holds, for an open subset $U$ of $X$ consider the map $X\rightarrow \{ \bullet_u \rightarrow \star_{a=b} \leftarrow \bullet_{v} \}$ where $U$ is the preimage of $\bullet_u$, and $X\setminus \bar{U}$ is the preimage of $\bullet_v$. Continuity of the lifting is then equivalent to the closure of $U$ being open.

Sorry I missed that "generator" and "weak generator" are the same thing defined in Definition 13.36.3. In this situation, when you say "$E$ generates $\mathcal{D}$ " do you mean that $E$ is a weak generator of $\mathcal{D}$ ?

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"Let $\mathcal{F}$ be a presheaf $\mathcal{O}$-modules" should be "Let $\mathcal{F}$ be a presheaf of $\mathcal{O}$-modules".

A search using either "generator" or "generators" did not show the definition of generator before this Lemma. By "generator" do you mean the generator defined in Section 19.10 ?

Sorry, but now I cannot understand which lemma this refers to. Please use tags when referring to things OR (preferable) comment on the page of the tag you are talking about.

Thanks to you both. Fixed here.

Thanks, fixed here.

Indeed. FIxed here.

Oops. Added here.

OK, I changed this here.

Thanks. Fixed here.

Going to leave as is.

The last ) in the proof I think is redundant

Consider the following map of finite topological spaces, hopefully in suggestive notation: $\pi: \{ \bullet_u \rightarrow \star_{a=b} \} \uplus \{ \bullet_{v} \rightarrow \star_{b} \} \longrightarrow \{ \bullet_u \rightarrow \star_{a=b} \leftarrow \bullet_{v} \}$

Here $\uplus$ denotes the disjoint union, the subscripts indicate where each point goes, and $\bullet_u \rightarrow \star_a$ means that point $\star_a$ lies in the closure of point $\bullet_u$.

Then we can add the following item to Proposition 5.26.6(08YN). I state it in full:

Proposition 5.26.6' (08YN'). Let X be a topological space. The following are equivalent

X is extremally disconnected,


for any solid commutative diagram $X \longrightarrow Z_0 \xleftarrow \pi Y_0$
of continuous maps with the surjective proper map $\pi:Y_0\longrightarrow Z_0$ of finite topological spaces defined above,  there is a dotted arrow $X\longrightarrow Y_0$ in the category of topological spaces making the diagram commute.

Proof. It is clear that (1) implies (2). On the other hand, if (2) holds, for an open subset $U$ of $X$ consider the map $X\rightarrow \{ \bullet_u \rightarrow \star_{a=b} \leftarrow \bullet_{v} \}$ where $U$ is the preimage of $\bullet_u$, and $X\setminus \bar{U}$ is the preimage of $\bullet_v$. Continuity of the lifting is then equivalent to the closure of $U$ being open.

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Sure - I was just trying to say that Lemma 4.35.9 implies something slightly stronger, after the present lemma is proved. And as for the reference to Lemma 100.8.4, I think I was trying to say that the proof idea in loc. cit. would give a shorter proof of the present lemma (though I have since forgotten whether the proofs might be mostly the same anyways).

(Feel free to ignore this comment.)

Going to leave as is.

Thanks Elias. Going to leave this as is for now.