Are category of groups and abelian groups topoi?

Question

I am currently learning Grothendieck topoi and am wondering if Grps and Abel are topoi. I guess they are not topoi but I do not know which one of the Giraud axiom fails. The literature I am using is Categorical logic written by Jacob Lurie. According to Lurie there are six Giraud Axioms:

G1: The category $\mathscr{X}$ admits finite limits.
G2: Every equivalence relation in $\mathscr{X}$ is effective.
G3: The category $\mathscr{X}$ admits finite coproducts, and coproducts are disjoint.
G4: The collection of effective epimorphisms in $\mathscr{X}$ are closed under pullback.
G5: Coproducts commute with pullbacks.
G6: There exists a set of objects $\mathscr{U}$ of $\mathscr{X}$ which generate $\mathscr{X}$ in the following sense: for every object $X\in \mathscr{X}$, there exists a covering {$U_i\rightarrow X$}, where each $U_i$ belongs to $\mathscr{U}$.

Abel satisfies G1 and G3. The main task is to consider if it satisfies the rest. Let us consider Abel first:
My main idea is to consider the adjunction$$\text{Hom}_{Abel}(\mathbb{Z}[S],G)\simeq \text{Hom}_{Sets} (S,UG)$$ Here $\mathbb{Z}[S]:=\bigoplus_{s\in S}\mathbb{Z}$ and $U$ denotes forgetful functor. Note that in such case forgetful functor $U$ is faithful.

I think Abel satisfy G4

$\require{AMScd}$ Let\begin{CD} G\times_H K @>>> K\\ @VVV @VVV\\ G @>>> H \end{CD}be a pullback square in Abel such that $G\rightarrow H$ is effective epi. We want to show $G\times_{H}K \rightarrow K $ is effective epimorphism. That is, if we have the following coequalizer digram $$(G\times_H K)\times_K (G\times_H K)\rightarrow (G\times_H K)\rightarrow W$$ then the induced morphism $W\rightarrow K$ is isomorphism( We have this morphism because $K$ also coequalizes the maps. ). We know in Abel, epimorphisms are the same thing as surjection. Hence, after applying $U$ we get epimorphism in $Sets$ and in $Sets$ epimorphisms are effective. And we know $Set$ is a topos. Therefore, by Giraud axioms top map of the following square is effective \begin{CD} UG\times_{UH} UK @>>> UK\\ @VVV @VVV\\ UG @>>> UH \end{CD} This means if we have the following coequalizer diagram in Sets $$(UG\times_{UH} UK)\times_{UK} (UG\times_{UH} UK)\rightarrow (UG\times_{UH} UK)\rightarrow UW$$ then the induced morphism $UW\rightarrow UK$ is isomorphism in Sets. Note that $U$ is right adjoint, which does not necessarily preserves colimits but I think it should preserve coequalizers because in our situation the coequalizer in Abel is just set theoritic coequalizer. That is why this coequalizer diagram can be obatined by applying the forgetful functor $U$ to the coequalizer diagram in Abel above.
Finally, recall that $U$ is faithful. It can detect isomorphism, i.e. $\forall f, U(f) \text{ iso}\Rightarrow f\text{ iso}$. We know that $W\rightarrow K$ is isomorphism in Abel.

Definition If $X$ is an object of $\mathscr{X}$, an equivalence relation $R$ on $X$ is a monomorphism $R\rightarrow X\times X$ in $\mathscr{X}$ such that for any object $Y$, the image of the induced map $$\text{Hom}_{\mathscr{X}}(Y,R)\rightarrow \text{Hom}_{\mathscr{X}}(Y,X\times X)\simeq \text{Hom}_{\mathscr{X}}(Y,X)\times \text{Hom}_{\mathscr{X}}(Y,X)$$ is a equivalence relation on the set $\text{Hom}_{\mathscr{X}}(Y,X)$. An equivalence relation $R$ is called effective if $R\rightarrow X\times_{X/R}X$ is isomorphism. Here, $X/R$ means the coequalizer of $R\rightarrow X$.

I think Abel satisties G2.

Let $G\rightarrow H\times H$ be an equivalence relation in Abel. Let $G\rightarrow H\rightarrow Q$ be its associated coequalizer diagram, here $Q$ denotes the coequalizer. We want to show $G\rightarrow H\times_Q H$ is isomorphism.
I claim that $U$ preserves equivalence relation. Because $U$ is right adjoint, hence preserves limits and thus monomorphism. For the description of $\text{Hom}-$set we can consider the following diagram\begin{CD} \text{Hom}_{Set}(S,UG) @>>> \text{Hom}_{Set}(S,UH)\times \text{Hom}_{Set}(S,UH)\\ @VVV @VVV\\ \text{Hom}_{Abel}(\mathbb{Z}[S],G) @>>> \text{Hom}_{Abel}(\mathbb{Z}[S],H)\times \text{Hom}_{Abel}(\mathbb{Z}[S],H) \end{CD} where vertical morphism are iso by adjunction and bottom horizontal map is equivalence relation by assumption.
This means $UG\rightarrow UH\times UH$ is an equivalence relation in Sets. Using the fact that Sets is a topos we know that $UG\rightarrow UH\times UH$ is effective. This implies that $UG\rightarrow U(H\times_Q H)=UH\times{UQ} UH$ is isomorphism( Here I used again that forgetful functor preserves coequalizer. ). Using the fact that $U$ is faithful we know $G\rightarrow H\times H$ is effective in Abel.

I think Abel does not satisfies G5.

In Abel coproduct is just direct sum and product is direct product. Let us consider $G_1=G_2=\mathbb{Z}/3\mathbb{Z}$ and $H=\mathbb{Z}/2\mathbb{Z}$. Then $(G_1\oplus G_2)\times H$ contains $(3\times 3)\times 2=18$ elements but $(G_1\times H)\oplus (G_2\times H)$ contains $(3\times 2)\times (3\times 2)=6\times 6=36$ elements. Hence, they cannot be isomorphic.

I am confused about G6. G6 seems to be abstract to me and I do not know if Abel satisfies G6.

For Grps my initial thought was to consider the adjunction $$\text{Hom}_{Grps}(\ast_{s\in S}\mathbb{Z},G)=\text{Hom}_{Sets}(S,UG)$$ the forgetful functor is still faithful right adjoint but the problem is it does not preserves coequalizer. What can we say about Grps? Are G2 G4 still satisfied? Is my proof for Abel correct? Thanks :)

Answer 1

You are working way too hard! A topos must be cartesian closed, but $\text{Grp}$ and $\text{Ab}$ both have zero objects.

Proposition: If a cartesian closed category $C$ has a zero object then it is contractible (every pair of objects has a unique morphism between them; equivalently, $C$ is equivalent to the terminal category).

Proof. In a cartesian closed category $C$, $\text{Hom}(X, Y)$ satisfies

$$\text{Hom}(1, Y^X) \cong \text{Hom}(X, Y)$$

where $Y^X$ is the exponential and $1$ is the terminal object. If $C$ has a zero object then $1$ is the zero object so there is a unique morphism $1 \to Y^X$, namely the zero morphism. So there is a unique morphism $X \to Y$ between any two objects, namely the zero morphism. $\Box$

In terms of Giraud's axioms, I haven't checked carefully but you should find that both categories fail G5. Coproducts do not commute with pullbacks. G5 is a stronger version of being a distributive category, meaning a category with finite products and coproducts where finite products distribute over finite coproducts, and we also have:

Proposition: If a distributive category has a zero object then it is contractible.

Proof. In a distributive category, finite products distribute over the empty coproduct, meaning that

$$X \times 0 \cong 0$$

where $0$ is the initial object. And in any category with finite products we have

$$X \times 1 \cong X$$

where $1$ is the terminal object. In a category with zero objects, $0 = 1$, so combining these two isomorphisms gives $X \cong 0$, so every object is isomorphic to the zero object. $\Box$

Answer 2

Qiaochu Yuan’s answer gives a quick way to see that $\newcommand{\Grp}{\mathbf{Grp}}\Grp$ and $\newcommand{\Ab}{\mathbf{Ab}}\Ab$ can’t be toposes — but it’s also helpful to look very explicitly at how G5, “coproducts preserved by pullback”, fails.

Products are a special case of pullbacks, $A \times B \newcommand{\iso}{\cong}= A \times_1 B$, and so G5 implies directly that product distributes over coproducts, $A \times \coprod_i B_i \iso \coprod_i (A \times B_i)$. In particular, the nullary and binary cases say that $A \times 0 \iso 0$, and $A \times (B_1 + B_2) \iso (A \times B_1) + (A \times B_2)$.

But in $\Ab$ and $\Grp$, these fail for non-trivial $A$ and any $B_i$:

• $A \times 0 \iso A$, since $0$ is terminal;
• in $\Ab$, $A \times (B_1 + B_2) \iso A \times B_1 \times B_2$, since finite products and coproducts coincide;
• in $\Grp$, the coproduct is the free product $X \ast Y$; so in $(A \times B_1) \ast (A \times B_2)$ the two natural inclusions of $A$ are distinct for $A$ nontrivial, whereas in $A \times (B_1 \ast B_2)$ there is just one natural such inclusion.

One possible way to track down this as the point where the Giraud axioms fail for $\Grp$ and $\Ab$ is to first notice that they are not cartesian closed (as Qiaochu’s answer starts with), and then notice that the Giraud axioms most closely connected to cartesian closure are G4 and G5, since “pullback is a left adjoint” is closely connected to “pullback preserves colimits”. I like to speak of the “moral adjoint functor theorem”, the statement “a functor is a left (resp. right) adjoint iff it preserves all colimits (resp. limits)” — this is not quite true, and the various actual adjoint functor theorems fix it, but it’s a very useful heuristic.