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Let $X$ be a topological space and $\mathcal{F}$ be a sheaf of sets. Let $\mathrm{Spe}({\mathcal{F}}):=\bigsqcup_{x\in X}\mathcal{F}_x$ be the underlie set of the etale space of $\mathcal{F}$. We expect to equip it with a topology s.t.

  1. the projection map $p:\mathrm{Spe}({\mathcal{F}})\to X,s_x\mapsto x$ is continuous, i.e. $\bigsqcup_{x\in U}\mathcal{F}_x$ is open for all open $U\subset X$;
  2. $\forall U\in \mathrm{Top}(X),\forall s\in \mathcal{F}(U),s^\#:U\to \mathrm{Spe}({\mathcal{F}}),x\mapsto s_x$ is continuous.
  3. $\forall U\in \mathrm{Top}(X),$ for all set maps $f:U\to \mathrm{Spe}({\mathcal{F}})$ over $X$ that is not of the form $s^\#$ for some $s\in \mathcal{F}(U)$, $f$ is not continuous.

With all the three conditions holding, then the sheaf of sections associated to $p:\mathrm{Spe}({\mathcal{F}})\to X$ represents the sheaf $\mathcal{F}$. And the reverse is also true.

Clearly the topology $\tau$ generated by the subbase $\{s^\#(U):U\in\mathrm{Top}(X),s\in\mathcal{F}(U)\}$ satisfies all the conditions.

Update: By this link Final topology, it states the finest topology on $X$ s.t. a family of set maps $f_i:Y_i \to X$ with topological spaces $Y_i$ and set $X$, is continuous, uniquely exists. It can be shown that $\tau$ is the finest topology on $X$ s.t. condition 2 holds (also stated in Hartshorne, see Ex 1.13 on pg.67). Hence any topology satisfying all three conditions is contained in $\tau$.

The question is:

if there exists another topology $\tau^\prime$ satisfying the above three conditions, do we must have $\tau =\tau^\prime$?

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  • $\begingroup$ $Spe(\mathcal F)$ carries the weakest topology with respect to $p$, which means that if the projection map is also continuous with respect to another topology $\tau'$, then $\tau\subseteq \tau'$. Since $\tau$ is also the strongest according to Hartshorne, we must have $\tau'\subseteq \tau$, which implies that $\tau = \tau'$. Notice however that I have only used conditions (1) and (2) to prove the uniqueness. Is it possible to derive (3) from the other 2? $\endgroup$ Commented yesterday
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    $\begingroup$ @XavierO the topology $\tau$ is not the weakeast topology w.r.t. to $p$. The weakeast topology w.r.t. $p$ only consists of open sets of the form $\bigsqcup_{x\in U}\mathcal{F}_x$ with open $U\subset X$. $\endgroup$ Commented yesterday
  • $\begingroup$ Oh yes, sorry. I got it wrong. Although it still holds that $\tau'\subset eq \tau$ from Hartshorne. $\endgroup$ Commented yesterday
  • $\begingroup$ @XavierO That's right, I got that part, now it remains to show uniqueness from the three conditions. $\endgroup$ Commented yesterday

1 Answer 1

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Let $X$ be a one-point space and $\mathcal{F}$ a sheaf on $X$, i.e. a choice of set $S=\mathcal{F}_{\ast}$. The condition 1. becomes entirely vacuous. The sheaf $\mathcal{F}$ has an empty section over the empty set and its sections over $X=\{\ast\}$ correspond to choices of elements in $S$, so every map $U\rightarrow\mathrm{Spe}(\mathcal{F})$ over $X$, $U\in\mathrm{Top}(X)$ is of the form $s^{\#}$ for some section $s$ of $\mathcal{F}$, i.e. condition $3$ is vacuous. Lastly, since $X$ is discrete, condition 2. is vacuous. Thus, any topology on $\mathrm{Spe}(\mathcal{F})=S$ satisfies conditions 1.-3.

In general, many different spaces $p\colon Y\rightarrow X$ can give rise to the same sheaf of sections. To get an equivalence, you want to consider those $p$ that are 'Γ©tale', i.e. local homeomorphisms. If you consider the condition 1'. that $p$ is a local homeomorphism, then you can show that $\tau$ satisfies 1'. and, in fact, $\tau$ is uniquely characterized by satisfying 1'. and 2. (the condition 3. is a consequence of those two).

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