points of a grothendieck topos $\mathscr{X}/X$

Question

There is an exercise related to Grothendieck topos that I cannot solve:

Let $e:\text{Set}\rightarrow \mathscr{X}$ be a point, i.e. a geometric morphism $e^*\dashv e_*:\text{Set}\rightarrow \mathscr{X}$. Then for each $X\in \mathscr{X}$ and $x\in e^*(X)$, there is a asscociated point of the topos $\mathscr{X}/X$. Give a description for the category of points of $\mathscr{X}/X$ where objects are given by pairs $(e,x)$ where $e$ is a point of $\mathscr{X}$ and $x\in e^*(X)$.

The following is what I think but probably wrong because it seems I constructed a point of $\mathscr{X}/X$ without touching elements in $e^*(X)$:

From the adjunction $e^*\dashv e_*$ we can get $\text{Hom}_{\text{Set}}(e^*(Y),S)=\text{Hom}_{\mathscr{X}}(Y,e_*(S))$. Now we can construct two functors $f^*:\mathscr{X}/X\rightarrow \text{Set}:(Y\rightarrow X)\mapsto e^*(Y)$ and $\text{Set}\rightarrow \mathscr{X}/X:S\mapsto (pr_X:(e_*(S)\times X)\rightarrow X)$. Now we claim $$\text{Hom}_{\text{Set}}(f^*(Y\rightarrow X),S)=\text{Hom}_{\mathscr{X}/X}((Y\rightarrow X),f_*(S))$$ Since $\text{Hom}_{\text{Set}}(f^*(Y\rightarrow X),S)=\text{Hom}_{\text{Set}}(e^*(Y),S)=\text{Hom}_{\mathscr{X}}(Y,e_*(S))= \text{Hom}_{\mathscr{X}/X}((Y\rightarrow X),(e_*(S)\times X\rightarrow X))=RHS$

The second last equation is due to that for any fixed morphism $g:Y\rightarrow X$ any morphism $h:Y\rightarrow e_*(S)\times X$ in $\mathscr{X}/X$ we have $pr_x\circ h=g$.

What is correct way to solve this exercise? Did I make any mistakes?

Answer 1

Your mistake is that your $f^\ast$ does not preserve finite limits.

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Think about it geometrically like this: Take a topos $\mathscr{X}$. We are supposed to view this as the category of sheaves on some space. Now take some $X \in \mathscr{X}$. This is a sheaf on $\mathscr{X}$, i.e. a space sitting via a local homeomorphism above $\mathscr{X}$. More precisely, there is a canonical geometric morphism $\newcommand\quotient[2]{{^{\Large #1}}/{_{ \Large #2}}} \smash{\quotient{\mathscr{X}}{X} \to \mathscr{X}}$. This consists of the following data:

• The functor $\require{AMScd} f^\ast: \mathscr{X} \to \quotient{\mathscr{X}}{X}, f^\ast(Y) = \left( \begin{CD} Y \times X \\ @VV\operatorname{pr}_XV \\ X \\ \end{CD} \right)$

• The right adjoint $f_\ast: \quotient{\mathscr{X}}{X} \to \mathscr{X}$. Note that this generally has no explicit description on sheaves.

• A further left adjoint $f_!: \quotient{\mathscr{X}}{X} \to \mathscr{X}, f_! \left( \begin{CD} Y \\ @VVV \\ X \\ \end{CD} \right) = Y$

with $f_! \dashv f^\ast \dashv f_\ast$. However, note that $f_!$ does not preserve finite limits. Only $f^\ast \dashv f_\ast$ gives a geometric morphism. $f_! \dashv f^\ast$ does not give a geometric morphism.

Now, what you're doing is the following: You're considering the compositions $$ \quotient{\mathscr{X}}{X} \overset{f_!}{\to} \mathscr{X} \overset{e^\ast}{\to} \mathsf{Set} $$ and $$ \mathsf{Set} \overset{e_\ast}{\to} \mathscr{X} \overset{f^\ast}{\to} \quotient{\mathscr{X}}{X} $$ Now, clearly, $e^\ast \circ f_! \dashv f^\ast \circ e_\ast$, but there is no reason for $e^\ast \circ f_!$ to preserve finite limits.

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In fact, here is a concrete example: Take $\mathscr{X} = \mathsf{Set}$ and $X = 2$. Then $\smash{\quotient{\mathsf{Set}}{2} = \mathsf{Set}^2}$. The canonical morphism $\mathsf{Set}^2 \to \mathsf{Set}$ is geometrically just the projection $2 \to 1$. Now we have that $f_!(S, T) = S \sqcup T$, $f^\ast(S) = (S, S)$ and $f_\ast(S, T) = S \times T$. Now it is clear that $f_!$ does not preserve products, as for example $$ 2 = f_!(1, 1) = f_!((1, 1) \times (1, 1)) \neq f_!(1, 1) \times f_!(1, 1) = 2 \times 2 = 4 $$

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The situation is quite analogous to commutative algebra: There we look at $K$-algebras $A$, and we are supposed to view these as rings of functions on some space. A point of $A$ is just a $K$-algebra homomorphism $A \to K$. But then also the map $K^2 \to K, (x, y) \mapsto x + y$ is a morphism of $K$-vector spaces (corresponding to the functor $f_!$ preserving colimits), but it does not preserve products (corresponding to the functor $f_!$ not preserving finite limits), so it is not a $K$-algebra homomorphism (corrsponding to the adjoint pair $f_! \dashv f^\ast$ not being a geometric morphism). This is because geometrically, ring homomorphisms $K^2 \to K$ correspond to maps $1 \to 2$, and of course there are exactly two such maps (the inclusions of the two points), corresponding to the $K$-algebra homomorphisms $K^2 \to K, (x, y) \mapsto x$ and $K^2 \to K, (x, y) \mapsto y$.

Answer 2

You're right to worry about your attempt, as Smiley1000 shows.

Instead, let's follow the hint. We're told that a point of $\mathscr{X}/X$ should be a pair $(e,x)$ where $e : \mathsf{Set} \to \mathscr{X}$ is a point of $\mathscr{X}$ and $x \in e^*(X)$. To intuitively see why, remember we can think of $\mathscr{X}/X$ as being a kind of "etale space" over $\mathscr{X}$. For simplicity let's picture a covering space $p : \mathscr{Y} \to \mathscr{X}$ -- then a point of $\mathscr{Y}$ is the data of a point in $\mathscr{X}$ and the choice of lift in the stalk $\mathscr{Y}_x = p^{-1}(x)$... I'll leave it as a nice exercise to see how this definition for spaces translates along the inclusion of spaces into topoi to become the hint.

So we want $(e,x)$ to somehow give us an adjunction between $\mathsf{Set}$ and $\mathscr{X}/X$... Let's do the inverse image first. Again, we think of an object $Y \in \mathscr{X}/X$ (really the topos $(\mathscr{X}/X)/Y$) as being an etale space over $\mathscr{X}/X$ whose fibre at a point $(e,x)$ is exactly $(e,x)^*(Y)$. But of course, $(\mathscr{X}/X)/Y \to \mathscr{X}/X \to \mathscr{X}$ means this is a covering space of $\mathscr{X}$ too, and we can relate these two perspectives! Indeed, some meditation on covering spaces shows that a point in the fibre at $(e,x)$ is just a point in the fibre over $e$ which maps to $x$. Again, translating this into the language of topoi we learn that the inverse image will be

$$ (e,x)^*(p: Y \to X) = \{ y \in e^*(Y) \mid e^*(p)(y) = x \} $$

(using the fact that $e^*$ is a functor, so that $e^*(p) : e^*(Y) \to e^*(X)$). I'll leave it to you to check that this is cocontinuous and finitely continuous.

Now the question asks about the whole category of points, so we want to understand maps between these... Well, a map $\Phi : e \to f$ is just a natural transformation $e^* \to f^*$. We would like to build a natural transformation $(e_1,x_1)^* \to (e_2,x_2)^*$. So we want, for every $p : Y \to X$, a map from $(e_1,x_1)^*(p) \to (e_2,x_2)^*(p)$ which is natural in $p$. Explicitly, this is

$$ \{y_1 \in e_1^*(Y) \mid e_1^*(p)(y) = x_1 \} \to \{y_2 \in e_2^*(Y) \mid e_2^*(p)(y) = x_2 \} $$

We have access to all our natural transformations $\Phi : e_1^* \to e_2^*$, and we know $\Phi_Y : e_1^*(Y) \to e_2^*(Y)$, so we're most of the way there! If moreoever $\Phi_X(x_1) = x_2$ then naturality of $\Phi$ will tell us that

$$e_2^*(p)(\Phi_Y(y_1)) = \Phi_X(e_1^*(p)(y_1)) = x_2$$

so that whenever $\Phi_X(x_1) = x_2$, $\Phi$ restricts to a natural transformation $(e_1,x_1)^* \to (e_2,x_2)^*$!

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So where did we end? We learned that the points of $\mathscr{X}/X$ are pairs $(e,x)$ where $e$ is a point of $\mathscr{X}$ and $x \in e^*(X)$. Moreover, the inverse image part of $(e,x)$ sends an object $p : Y \to X$ to the set $\{y \in e^*(Y) \mid e^*(p)(y) = x \}$. The natural transformations $(e_1,x_1)^* \to (e_2,x_2)^*$ are exactly those natural transformations $\Phi : e_1^* \to e_2^*$ with the ~bonus property~ that $\Phi_X(x_1) = x_2$.

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I hope this helps ^_^