If $X$ is a sober topological space, the real numbers object in the topos $\mathrm{Sh}(X)$ is the sheaf of continuous real-valued functions on $X$. This is proven very explicitly in Theorem VI.8.2 of MacLane & Moerdijk Sheaves in Geometry and Logic by compiling out the definition of real numbers in Kripke-Joyal semantics. A more abstract proof is in D4.7.6 of Sketches of an Elephant: since $\mathrm{Sh}(\mathbb{R})$ is the classifying topos of the theory of a real number, maps $U \to R_D$ in $\mathrm{Sh}(X)$ from an open subset $U\subseteq X$ to the real numbers object are equivalent to geometric morphisms $\mathrm{Sh}(X)/U \to \mathrm{Sh}(\mathbb{R})$, but $\mathrm{Sh}(X)/U \simeq \mathrm{Sh}(U)$ and so these are equivalent to continuous maps $U\to \mathbb{R}$.

Theorem VI.9.2 of MacLane & Moerdijk makes an analogous claim for $\mathrm{Sh}(\mathbf{T})$, where $\mathbf{T}$ is a full subcategory of topological spaces closed under finite limits and open subspaces. My question is about a glossed-over point in the proof: they construct maps back and forth between sections of $R_D$ and the continuous $\mathbb{R}$-valued functions, but they don't say anything about why the maps are inverses. I believe it in the situation of $\mathrm{Sh}(X)$, but for $\mathrm{Sh}(\mathbf{T})$ it's not obvious to me, because a continuous map $Y\to \mathbb{R}$ only "knows" about the open subspaces of $Y$, whereas a map $Y \to R_D$ in $\mathrm{Sh}(\mathbf{T})$ says something about all continous maps $Z\to Y$ in $\mathbf{T}$. The most I can see how to show is that the sheaf of continuous real-valued functions is a retract of $R_D$.

In terms of the abstract proof from the Elephant, the question is this: for $Y\in \mathbf{T}$, geometric morphisms $\mathrm{Sh}(\mathbf{T})/Y \to \mathrm{Sh}(\mathbb{R})$ are equivalent to continuous real-valued functions on the locale corresponding to the frame of subterminals in $\mathrm{Sh}(\mathbf{T})/Y$, but this frame is not (as far as I can see) the same as the open-set frame of $Y$.

  • 1
    $\begingroup$ Maybe I'm remembering the wrong thing, but I recall the argument of MacLane & Moerdijk as first proving that if $R$ is the real numbers object, that $R(Y)$ was the set of continuous maps $Y \to R$. From this, it follows that $\mathbb{R}$ is a representing object for this sheaf, and by Yoneda's lemma, $R \cong \mathbf{y}\mathbb{R}$. $\endgroup$
    – user13113
    Nov 1, 2014 at 0:25
  • 1
    $\begingroup$ @Hurkyl yes, that's exactly how it goes. My question is about the first part of the argument. $\endgroup$ Nov 2, 2014 at 4:25
  • 1
    $\begingroup$ In your introductory paragraph, why do you need to assume that $X$ is sober? Sobriety seems superfluous to me. $\endgroup$ Nov 4, 2014 at 15:21
  • 1
    $\begingroup$ @IngoBlechschmidt you're probably right. $\endgroup$ Nov 4, 2014 at 21:06
  • 8
    $\begingroup$ @IngoBlechschmidt, I know that it's nearly 4 years later, but I just love the sentence "Sobriety seems superfluous to me." $\endgroup$
    – LSpice
    May 7, 2018 at 17:44

1 Answer 1


Following a suggestion of Thomas Holder, we can close the gap as follows:

  1. For each object $Y$ in $\mathbf{T}$, there is a pseudonatural local geometric morphism $\mathbf{Sh}(\mathbf{T}_{/ Y}) \to \mathbf{Sh} (Y)$.
  2. $\mathbb{R}$ is Hausdorff, so for any Grothendieck topos $\mathcal{E}$, the category of geometric morphisms $\mathcal{E} \to \mathbf{Sh} (\mathbb{R})$ is essentially discrete.
  3. Essentially by definition, any local geometric morphism is a right adjoint in the bicategory of toposes. Thus, we have a pseudonatural left adjoint $$\mathbf{Geom} (\mathbf{Sh} (Y), \mathbf{Sh} (\mathbb{R})) \to \mathbf{Geom} (\mathbf{Sh} (\mathbf{T}_{/ Y}), \mathbf{Sh} (\mathbb{R}))$$ but any adjunction between groupoids is an equivalence, so we deduce that $\mathbf{T}(-, \mathbb{R})$ is not merely a retract of $R_D$ in $\mathbf{Sh} (\mathbf{T})$ but actually isomorphic to $R_D$, at least when every object in $\mathbf{T}$ is sober.
  • $\begingroup$ Where does sobriety of the objects of T enter? $\endgroup$ Nov 4, 2014 at 21:08
  • $\begingroup$ Don't we need it to deduce that $\mathbf{Geom}(\mathbf{Sh}(Y), \mathbf{Sh}(\mathbb{R}))$ is equivalent to the set $\mathbf{T}(Y, \mathbb{R})$? $\endgroup$
    – Zhen Lin
    Nov 4, 2014 at 23:12
  • $\begingroup$ But $\mathbb{R}$ is sober, so any map $Y\to \mathbb{R}$ factors through the soberification of $Y$? $\endgroup$ Nov 5, 2014 at 5:50
  • 2
    $\begingroup$ Just to confirm, I agree that we don't need sobriety. Without any condition on $Y$, the following notions coincide: (1) continuous maps from $Y$ to the topological space of Dedekind reals, (2) locale-theoretic morphisms from $Y$, now considered as a locale, to the locale of Dedekind reals, (3) geometric morphisms from $\mathbf{Sh}(Y)$ to the topos of Dedekind reals. (If classical logic is available, then this topos coincides with the topos of sheaves over the topological space of Dedekind reals. If not, we can use the classifying topos of Dedekind reals as a substitute for that sheaf topos.) $\endgroup$ May 17, 2019 at 1:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.