# Does the internal axiom of choice imply Lagrange's theorem?

In a topos ${\mathcal{A}}$, given a group object $G$ and a subgroup $H$, the object $H\backslash G$ of right cosets is the coequalizer of two maps $G\times H\rightrightarrows G$, namely the group multiplication and the projection onto $G$. Denote the coequalizing map by $H\backslash -$: $$G\times H\rightrightarrows G\xrightarrow{H\backslash -}H\backslash G$$

Say that a map $f:X\to Y$ is a product projection if it is part of a span $Z\leftarrow X\xrightarrow{f} Y$ that writes $X$ as a product of $Y$ and $Z$. One can show that $H\backslash -:G\to H\backslash G$ has a section if and only if it is a product projection. If so, then in fact $G$ is the product of $H$ and $H\backslash G$: $$G\simeq H\times H\backslash G$$

In the topos ${\mathcal{FinSet}}$ of finite sets, the external axiom of choice holds, which is to say that every epimorphism has a section. The map $H\backslash -$ is an epimorphism, hence has a section if $G$ is a finite group, hence $G\simeq H\times H\backslash G$. This is (a strong form of) Lagrange's theorem.

Say that a topos ${\mathcal{A}}$ satisfies Lagrange's theorem if, for each group object $G$ and a subgroup $H$, the map $H\backslash -:G\to H\backslash G$ has a section (or equivalently, is a product projection). For example, the topos ${\mathcal{Top}}$ of topological spaces and continuous maps does not satisfy Lagrange's theorem as the Hopf fibration $S^1\to S^3\to S^2$ is a counterexample.

The external axiom of choice implies Lagrange's theorem. Hence the topos ${\mathcal{Set}}$ satisfies Lagrange's theorem. The internal axiom of choice, namely that exponentials $-^X:{\mathcal{A}}\to{\mathcal{A}}$ preserve epimorphisms, is weaker than the external axiom of choice. My question is:

Does there exist a topos that satisfies the internal axiom of choice but not Lagrange's theorem?

• The category $\mathbf{Top}$ is not a topos: it is not cartesian closed, it has no subobject classifier, etc. etc. I'm also inclined to say your formulation of Lagrange's theorem is wrong: even in $\mathbf{Set}$, the bijection $G \cong (G / H) \times H$ is highly non-canonical. Oct 29 '13 at 18:03
• There exists a product decomposition of $G$ into $H\times H\backslash G$ for each section of $G \to H\backslash G$.
– user2529
Oct 29 '13 at 19:20

Consider the topos $\mathcal E$ of $\mathbb Z/2$-sets. It satisfies the internal axiom of choice but not the external one. The Klein four-group $(\mathbb Z/2)\times(\mathbb Z/2)$ admits an action of $\mathbb Z/2$, namely interchanging the two factors, which makes it an object of $\mathcal E$. Let $H$ be the diagonal subgroup, $\{(0,0),(1,1)\}$ and note that it is pointwise fixed by the $\mathbb Z/2$-action. The quotient $H\backslash G$ consists of two points, both of which are fixed by the $\mathbb Z/2$-action. Therefore, the product $H\times(H\backslash G)$ consists of four points, each fixed by the $\mathbb Z/2$-action. But $G$ has only two fixed-points, the other two points, namely $(0,1)$ and $(1,0)$, being interchanged by the action.
• Why is $H$ a subobject of the Klein four group? It seems that $H$ is not closed under the action of interchanging the two factors of the Klein four group.
• There is no global element $1\to Z/2 \times Z/2$ of the latter object with the interchanging action. The obstruction to finding such a equivariant morphism is due to the action on the final object $1$ being trivial while the action on $Z/2 \times Z/2$ is free. Hence $Z/2 \times Z/2$ is only a semigroup object and not a group object.
• @ColinTan: Interchanging the two factors in the Klein 4-group sends $(0,0)$ and $(1,1)$ to themselves and sends $(0,1)$ and $(1,0)$ to each other. In particular, contrary to what you assert, the action on $Z/2\times Z/2$ is not free. $H$ is not only closed under the action but pointwise fixed. (Note that, in any group object in any topos, the identity element of the group is always a global element.) Oct 31 '13 at 18:22