math.kleen.org

Hi there! My name is Viktor Kleen. I’m currently a PhD candidate in Mathematics at USC in Los Angeles, see my CV. My research interests include:

$\AA^1$ -homotopy theory
Higher category theory
Algebraic geometry

I now have a preprint describing my work on a motivic version of Snaith’s stable splitting results.

This site is supposed to serve as a research notebook to get me to write up ideas and results. As such, it is woefully unpolished, unreliable and generally to be used at your own risk.

Still, have fun! And please send email with suggestions, corrections, criticism or general rants to kleen@usc.edu if you like.

Torsors and Their Classification III

Interlude on constructions with torsors

Let $P$ be a right torsor for some group object $G$ . Twisting with $P$ will be a functor $\objl{G}\to \ca C$ from objects with a left $G$ –action to objects in $\ca C$ .

Let $X$ be an object with a left $G$ –action. The twist $P\times^G X$ is defined as the quotient of $P\times X$ by the diagonal action of $G$ , that is, it is the coequalizer $\begin{tikzcd} G\times P\times X \arrow[r,shift left=.5ex, "\act"] \arrow[r, shift right=.5ex, "\proj"'] & P\times X \arrow[onto,r] & P\times^G X \end{tikzcd}$ where $\act(g, p, x) = (p g^{-1}, gx)$ on generalized elements. If $X$ additionally carries a right action by some group object $H$ , then evidently the twist $P\times^G X$ also carries a right $H$ –action; similarly, if the torsor $P$ carries a left action by some group object $H'$ , then the twist also carries a left $H'$ –action.

Let’s introduce some notation for the action of the epimorphism $P\times X\onto P\times^G X$ on generalized elements. Namely, given a generalized element $(p,x)\colon T\to P\times X$ , I will write $[p,x]$ for the composition $T\to P\times X\to P\times^G X$ . Because $P\times X\onto P\times^G X$ is an epimorphism, to check some relation between generalized elements downstairs it will be enough to check it on generalized elements of the form $[p,x]$ . Furthermore, if $(p,x)$ actually comes from a generalized element $(g,p,x)\colon T\to G\times P\times X$ via projection, then by the coequalizer property we find $[p,x] = [p g^{-1}, g x]$ .

The process of twisting has some exactness properties:

The functor $P\times^G{-}\colon\objl{G}\to\ca C$ preserves colimits and finite products.

Regarding preservation of colimits, we can in fact construct a right adjoint to $P\times^G{-}$ . It is given by the functor $({-})^P$ which sends $T$ to the function object $T^P$ whose $G$ –action is given by precomposing with the action on $P$ . This won’t be particularly important in what follows, so I’ll skip the detailed calculation.

Showing that binary products are preserved is a little more difficult. Let’s first observe that the functor $G\times^G{-}$ is naturally isomorphic to the identity functor: Consider the commutative diagram $\begin{tikzcd} G\times X & X \arrow[l, "e\times{\id}"'] \\ G\times G\times X \arrow[u, shift left=.5ex, "\proj"] \arrow[u, shift right=.5ex, "\act"'] & X \arrow[l, "e\times e\times{\id}"] \ar[u, shift left=.5ex, "{\id}"] \ar[u, shift right=.5ex, "{\id}"'] \end{tikzcd}$ This induces a natural transformation ${\id}\to G\times^G{-}$ which is easily checked to be an isomorphism on generalized elements.

Now, choose a trivializing cover $U\onto *$ of the torsor $P$ . Because pullback in a topos commutes with colimits and limits, it will then be enough to check that, for any $X$ and $Y$ , the natural morphism $P\times^G (X\times Y)\to (P\times^G X) \times (P\times^G Y)$ after pulling back to the corresponding natural morphism in $\ca C/U$ becomes an isomorphism. Since there is a $G\times U$ –equivariant isomorphism $P\times U\isom G\times U$ , the functors $P\times U\times^{G\times U}{-}$ and $G\times U\times^{G\times U}{-} \isom \id$ are naturally isomorphic; in particular $P\times U\times^{G\times U}{-}$ preserves products, which is exactly what we needed to show.

Preservation of finite products in particular implies that $P\times^G{-}$ sends group objects in $\objl{G}$ to group objects in $\ca C$ . This allows the following construction.

When endowed with the conjugation action of $G$ on itself, $G$ becomes a group object in $\objl{G}$ . In particular, ${}^P G\coloneqq P\times^G G$ is naturally a group object in $\ca C$ .

It is enough to check that the structure maps $e\colon *\to G$ , $m\colon G\times G\to G$ and $i\colon G\to G$ are equivariant with respect to conjugation. This can be checked on generalized elements where it reduces to the corresponding statement about genuine groups. There, it is a simple calculation.

More generally, if $G$ acts on itself via automorphisms, then the twist $P\times^G G$ with respect to this action becomes a new group object. Sometimes, this is called a form of $G$ . This construction let’s us refine the process of twisting a little bit. It turns out that the twisted object $P\times^G X$ comes naturally equipped with a left group action by ${}^P G$ . In fact, $P$ itself carries a natural left action by ${}^P G$ . Recall that, since $P$ is a righ $G$ –torsor, the map $P\times G\to P\times P$ given by $(p,g)\mapsto (p, pg)$ on generalized elements is an isomorphism. Let’s denote the inverse map by $(p, q) \mapsto (p, \tdiv{p}{q})$ on generalized elements. Then we can construct a commutative diagram $\begin{tikzcd} P\times G \times P\arrow[r, "\varphi"] & P \arrow[d, equal] \\ G\times P\times G\times P\arrow[u, shift left=.5ex]\arrow[u, shift right=.5ex] \arrow[r, "\psi"'] & P \end{tikzcd}$ where, on generalized elements, $\varphi(p,g,q) = p\cdot g \tdiv{p}{q}$ and $\psi(h,p,g,q) = \varphi(p,g,q)$ . The commutativity follows from the following calculation: $\varphi(ph^{-1}, hgh^{-1}, q) = ph^{-1}\cdot hgh^{-1} \tdiv{ph^{-1}}{q} = p\cdot g h^{-1} h \tdiv{p}{q} = \varphi(p,g,q).$ Hence, because colimits commute with products in toposes, we have a morphism ${}^P G\times P\to P$ where the axioms for a left action can be checked by direct calculation.

This action makes $P$ into a left ${}^P G$ –torsor.

Take an epimorphism $U\onto *$ which trivializes $P$ . Then it will be enough to check that there is a ${}^P G$ –equivariant isomorphism $P\times U\iso {}^P G\times U$ . Since everything in sight commutes with pullbacks, we might as well assume that $P$ is isomorphic to the trivial right $G$ –torsor and check that then it is ${}^P G$ –isomorphic to the trivial left ${}^P G$ –torsor.

We first need to produce an isomorphism $P\iso {}^P G$ given a $G$ –isomorphism $\varphi\colon P\iso G$ . I claim that the composition $G \to[g\mapsto(e,g)] G\times G\onto G\times^G G$ is an isomorphism. The inverse map is defined by the commutative¹ diagram $\begin{tikzcd} G\times G \arrow[r, "\conj"] & G \arrow[d, equal] \\ G\times G\times G \arrow[u, shift left=.5ex]\arrow[u, shift right=.5ex] \arrow[r, "\conj\circ\proj"'] & G \end{tikzcd}$ where $\conj$ denotes the conjugation action. Hence, we get a chain of isomorphisms² $\begin{tikzcd} P \arrow[r,"\varphi"] & G \arrow[r, "{g\mapsto(e,g)}"] & G\times^G G \arrow[r, "\varphi^{-1}_*"] & P\times^G G = {}^P G. \end{tikzcd}$ The last thing to check is that this isomorphism $P\iso {}^P G$ is in fact ${}^P G$ –equivariant, that is, we need to check that the diagram $\begin{tikzcd} {}^P G\times P \arrow[r, "\act"] \arrow[d] & P \arrow[d] \\ {}^P G\times G \arrow[d] & G \arrow[d] \\ {}^P G\times (G\times^G G) \arrow[d] & G\times^G G \arrow[d] \\ {}^P G\times {}^P G \arrow[r, "m"'] & {}^P G \end{tikzcd}$ commutes. On generalized elements, the top composition gives $[p, g], q \mapsto [\varphi^{-1}(e), \varphi(p) g \varphi(p)^{-1}\varphi(q)]$ while the bottom composition is $[p,q], q\mapsto [p, g\cdot\tdiv{\varphi^{-1}(e)}{p}\varphi(q)\tdiv{p}{\varphi^{-1}(e)}] = [p, g\varphi(p)^{-1}\varphi(q)\varphi(p)]$ But this is equal to $[\varphi^{-1}(e), \varphi(p) g\varphi(p)^{-1}\varphi(q)\varphi(p)\varphi(p)^{-1}]$ which proves the required commutativity.

A consequence of this theorem is that the twisting functor $P\times^G{-}$ may be considered as a functor $\objl{G}\to\objl{{}^P G}$ . Furthermore, if $Q$ is a left $G$ –torsor, then we can choose an epimorphism $U\onto *$ which trivializes both $Q$ and $P$ . Then, because twisting commutes with products, we find $(P\times^G Q)\times U\isom G\times U \times^{G\times U} G\times U\isom G\times U\isom {}^P G\times U$ similarly to the proof of the theorem. Hence, twisting restricts to a functor $P\times^G{-}\colon \Tors[G]\to\Tors[{}^P G]$ from the category of left $G$ torsors to the category of left ${}^P G$ torsors. Furthermore, if $Q$ is a left $K$ –torsor and a right $H$ –torsor, $P$ a left $H$ –torsor and a right $G$ –torsor then there is a natural isomorphism $Q\times^H (P\times^G{-})\isom (Q\times^H P)\times^G{-}$ which follows from the fact that colimits commute with colimits and with binary products in toposes.

If $P$ is a right $G$ –torsor and hence a left ${}^P G$ –torsor, let $\op{P}$ denote the right ${}^P G$ –torsor with action given by $p\cdot g = g^{-1}\cdot p$ on generalized elements. Then $\op{P}\times^{{}^P G} P$ is a trivial right $G$ –torsor. Furthermore, ${}^{\op{P}}({}^P G)\isom G$ as group objects and along this identification $P\times^G \op{P}$ is a trivial right ${}^P G$ –torsor. It follows that $P\times^G{-}\colon \Tors[G]\to\Tors[{}^P G]$ is an equivalence of categories with weak inverse $\op{P}\times^{{}^P G}{-}\colon\Tors[{}^P G]\to\Tors[G].$

Let’s first produce an isomorphism ${}^{\op{P}}({}^P G)\iso G$ . Let $\varphi\colon P\times P\times G\to G$ be such that $\varphi(p,q,g) = \tdiv{q}{p} \cdot g \cdot \tdiv{p}{q}$ on generalized elements. It is straightforward, albeit tedious, to check that $\varphi$ descends to a group homomorphism $\op{P}\times^{{}^P G} (P\times^G G) = {}^{\op{P}}({}^P G)\to G.$ I claim that this is an isomorphism. To verify this claim, it will be enough to pull back along an epimorphism $U\onto *$ which trivializes both the right $G$ –torsor $P$ and the right ${}^P G$ –torsor $\op{P}$ , that is, we can assume that $P$ is trivial.³ So let $\psi\colon G\iso P$ be a $G$ –equivariant isomorphism. We can then set $\widetilde\varphi\colon G\to {}^{\op{P}}({}^P G)$ to be the morphism such that $\widetilde\varphi(g) = [\psi(e), [\psi(e), g]]$ on generalized elements. Computing the compositions $\varphi\circ\widetilde\varphi$ and $\widetilde\varphi\circ\varphi$ on generalized elements, we find $\varphi(\widetilde\varphi(g)) = \varphi([\psi(e), [\psi(e), g]]) = g$ and $\widetilde\varphi(\varphi([p, [q, g]])) = [\psi(e), [\psi(e), \tdiv{q}{p}\cdot g\cdot \tdiv{p}{q}]].$ Noticing that $[p, [q, g]] = [p_0, [p_0, \tdiv{p_0}{p}]\, [p_0, \tdiv{q}{p_0}\cdot g\cdot \tdiv{p_0}{q}]\, [p_0, \tdiv{p}{p_0}]]$ where $p_0 = \psi(e)$ finishes the proof of ${}^{\op{P}}({}^P G)\isom G$ .

Next, let’s check that $\op{P}\times^{{}^P G} P$ is a trivial right $G$ –torsor. Let $P\to P\times P\onto \op{P}\times^{{}^P G} P$ be the diagonal composed with the canonical projection. I claim that this map descends to the quotient $P/G\isom *$ which will provide a global section of $\op{P}\times^{{}^P G} P$ . This will then imply the claim. The calculation $[pg, pg] = [p, [p, g^{-1}]\cdot(pg)] = [p, p g g^{-1}] = [p,p]$ shows that the diagonal is $P\to \op{P}\times^{{}^P G} P$ is invariant under the right action of $G$ which proves the claim. The last assertion of the theorem is proven in a completely analogous way.

In summary, we have found that twisting with a right torsor $P$ is an invertible operation up to isomorphism. This will become essential in the computation of the homotopy groups of $BG$ at any basepoint.

Remember that in $G\times^G G$ the $G$ on the left is the trivial right torsor and the $G$ on the right is taken with the conjugation action.↩
Note that $\varphi^{-1}_*$ denotes the morphism induced by the $G$ –equivariant isomorphism $\varphi^{-1}\colon G\to P$ , using the fact that twisting is functorial with respect to $G$ –equivariant morphisms in both factors.↩
By the proof of the previous theorem, this also implies that $\op{P}$ is trivial.↩

Torsors and Their Classification II

Computing Homotopy Classes in Model Categories

Let again $\ca C$ be some Grothendieck topos and $G$ a group object in $\ca C$ . I eventually want to give a proof of the following theorem.

There is a functorial construction of a simplicial object $BG$ in $\ca C$ and a functorial bijection $\tors[G] \isom [*,BG]$ between the set of $G$ –torsors $\tors[G]$ and the set of simplicial homotopy classes of morphisms $*\to BG$ .

To show this we will follow an approach which was presented in the paper/preprint “Cocyle Categories” by Rick Jardine. The point is that proving our theorem comes down to computing homsets between objects of $\ca C$ in the homotopy category. Generally, in any model category $\ca M$ (and $\ca C$ embeds into the local model structure on the category ${\ca C}^{\op{\Simp}}$ of simplicial objects in $\ca C$ ) we can “calculate” homsets from $X$ to $Y$ in the homotopy category by taking a cofibrant replacement $X'$ of $X$ and a fibrant replacement $Y'$ of $Y$ and taking honest homotopy classes of maps from $X'$ to $Y'$ in $\ca M$ .

However, computing homsets like that tends to be quite difficult because, typically, cofibrant and fibrant replacement are hard to write down explicitly (at least in a form that one can work with). Jardine came up with the following work–around.

Let $\ca M$ be a category with weak equivalences and let $X$ and $Y$ be objects in $\ca M$ . Define a category $\coc(X,Y)$ of cocycles from $X$ to $Y$ whose objects are diagrams $\begin{tikzcd}[column sep=2ex, row sep=1ex] & Z \arrow[ld, "\sim"'{sloped,pos=-.1}] \arrow[rd] & \\ X & & Y \end{tikzcd}$ with $Z\iso X$ a weak equivalence and whose morphisms are the obvious commuting diagrams.

Remember that any category $\ca C$ has a nerve $\nerve(\ca C)\in\sset$ , its $n$ –simplices are composable chains of $n$ morphisms in $\ca C$ . By a slight abuse of notation I will not distinguish between $\coc(X,Y)$ and $\nerve(\coc(X,Y))$ . A slight remark regarding notation; the geometric realization of $\nerve(\ca C)$ is often called the classifying space of $\ca C$ and is denoted by $\class(\ca C)$ . Sometimes I will consider “a model of” $\class(\ca C)$ to be any convenient fibrant replacement of $\nerve(\ca C)$ in $\sset$ . In particular, it makes sense to talk about $\pi_0 \class(\ca C)$ and by another slight abuse of notation I will often just write $\pi_0(\ca C)$ .

It turns out that in any model category $\class(\coc(X,Y))$ is a good model for the mapping space from $X$ to $Y$ as long as $X$ is cofibrant and $Y$ is fibrant. This is proven for example in a preprint of Daniel Dugger’s, “Classification Spaces of Maps in Model Categories”. I will—following Jardine—give a proof of something weaker:

In a right proper model category $\ca M$ in which weak equivalences are preserved under products, the homset $[X,Y]$ in the homotopy category can in a natural way be computed as $[X,Y]\isom \pi_0\coc(X,Y)$

The map from right to left is easily described as $\begin{tikzcd}[column sep=2ex, row sep=1ex] & Z \arrow[ld, "f"'] \arrow[rd, "g"] & \\ X & & Y \end{tikzcd} \mapsto [g\circ f^{-1}]$ In the other direction, the map is harder to describe. Instead, let’s consider the set $\pi(X,Y)$ of (left) homotopy classes of maps from $X$ to $Y$ . Any map $X\to Y$ trivially defines a cocycle. If two such maps $f$ and $g$ are left homotopic, there is a cylinder object $\begin{tikzcd}X \arrow[r, yshift=3pt] \arrow[r, yshift=-3pt] & X\otimes I\end{tikzcd}$ and a homotopy $h\colon X\otimes I\to Y$ such that $\begin{tikzcd} & X \arrow[equal,dl] \arrow[d] \arrow[dr,"f"] & \\ X & X\otimes I\arrow[l, "\sim"'l] \arrow[r, "h" description] & Y \\ & X \arrow[equal,ul] \arrow[u] \arrow[ur, "g"'] & \end{tikzcd}$ This means that these three cocyles lie in the same path component of $\coc(X,Y)$ . Hence, we have a well defined map $\pi(X,Y)\to \pi_0\coc(X,Y)$ .

If $X$ is cofibrant and $Y$ is fibrant, then the map $\pi_0 \coc(X,Y)\to{} [X,Y]$ is a bijection.

To prove this, consider the commutative diagram $\begin{tikzcd} \pi(X,Y) \arrow[r] \arrow[dr, "\isom"'] & \pi_0\coc(X,Y) \arrow[d] \\ {} & {}[X,Y] \end{tikzcd}$ in which the diagonal map is a bijection because $X$ is cofibrant and $Y$ is fibrant. Hence, it suffices to show that our map is surjective, i.e. that any cocycle $(f,g)\colon \begin{tikzcd}[column sep=.8em] X & \arrow[l] Z \arrow[r] & Y\end{tikzcd}$ is in the path component of a cocycle $\begin{tikzcd}[column sep=.8em] X \arrow[equal,r] & X \arrow[r] & Y\end{tikzcd}$ . Factor $f$ into a trivial cofibration $j$ followed by a trivial fibration $p$ to obtain $\begin{tikzcd}[row sep=1ex] & Z \arrow[dl,"f"'] \arrow[dr,"g"] \arrow[dd, "j" description] & \\ X & & Y \\ & X' \arrow[ul, "p"] \arrow[dashed, ur, "\varphi"'] & \end{tikzcd}$ where the map $\varphi$ exists because $Y$ is assumed to be fibrant. Because $X$ is cofibrant we can find a section $\psi$ of $p$ and obtain a commutative diagram $\begin{tikzcd}[row sep=1ex] & X \arrow[dl,equal] \arrow[dr,"\psi\varphi"] \arrow[dd, "\psi" description] & \\ X & & Y \\ & X' \arrow[ul, "p"] \arrow[ur, "\varphi"'] & \end{tikzcd}$ which proves the lemma.¹

If $X'\iso X$ and $Y'\iso Y$ are weak equivalences, then the induced map $\pi_0\coc(X',Y')\to\pi_0\coc(X,Y)$ is a bijection.

To prove this we will describe an inverse function. Let $(f,g)\in\coc(X,Y)$ be a cocycle and think of it as a morphism $(f,g): Z\to X\times Y$ . Factor this map into a trivial cofibration followed by a fibration $\begin{tikzcd} Z \arrow[r, "j"] \arrow[dr, "{(f,g)}"'] & W \arrow[d, "{(p_X, p_Y)}"] \\ & X\times Y \end{tikzcd}$ and observe that by 2–out–of–3 $p_X$ is a weak equivalence. Form the pullback $\begin{tikzcd} W' \arrow[r, "\sim"] \arrow[d, "{(p_X^*,p_Y^*)}"'] & W\arrow[d, "{(p_X,p_Y)}"] \\ X'\times Y' \arrow[r, "\sim"] & X\times Y \end{tikzcd}$ in which the bottom map is a weak equivalence by our assumption on $\ca M$ and the top map is a weak equivalence by right properness. This then implies that $p_X^*$ is a weak equivalence by 2–out–of–3, i.e. $(p_X^*,p_Y^*)$ defines a cocycle in $\coc(X',Y')$ . This construction yields a well–defined map $\pi_0\coc(X,Y)\to\pi_0\coc(X',Y')$ which is inverse to the map in the lemma.

Using these two lemmas the theorem about computing homsets is easy to prove. Let $X$ and $Y$ be any two objects in $\ca M$ and let $X'$ be a cofibrant replacement of $X$ and $Y'$ a fibrant replacement of $Y$ . Then we have a commutative diagram $\begin{tikzcd} \pi_0\coc(X,Y)\ar[d, "\isom"'] \ar[r] & {}[X,Y] \ar[d, "\isom"] \\ \pi_0\coc(X',Y') \ar[r, "\isom"] & {}[X',Y'] \end{tikzcd}$ which proves the theorem.

Note that for this first lemma we haven’t used any assumption on the model category $\ca M$ .↩

Torsors and Their Classification I

Torsors in Grothendieck Toposes

Let $\ca C$ be some Grothendieck topos, that is a category of sheaves on some Grothendieck topology. This text will look at some properties of torsors in $\ca C$ and how they might be classified using cohomology and homotopy theory. For this we will first need to define group objects. Torsors will then be associated to those.

A group object in a category $\ca C$ is an object $G\in\ca C$ such that the associated Yoneda functor $\hom(\_,G)\colon \op{\ca C}\to\Set$ actually takes values in the category $\Grp$ of groups.

A trivial kind of example would be just a group considered as an object in $\Set$ . A more involved example would be a group scheme $G$ over some base $S$ . Such a thing is essentially defined as a group object in the category $\sch[S]$ . If we have any subcanonical topology on $\sch[S]$ , then $G$ defines a sheaf on the associated site and we obtain in this way a group object in the corresponding topos on $S$ .

Let’s now assume we have a group object $G$ in a Grothendieck topos $\ca C$ . Then we can define torsors over $G$ as follows:

A trivial $G$ –torsor is an object $X$ with a left $G$ –action which is isomorphic to $G$ itself with the action given by left multiplication.

A $G$ –torsor is an object $X\in\ca C$ with a left $G$ –action which is locally isomorphic to a trivial torsor; that is, there is an epimorphism $U\to *$ such that $U\times X$ is a trivial torsor in $\ca C/U$ over $U \times G$ .

I want to show that for any $G$ –torsor $X$ according to this definition the left action of $G$ on $X$ is free and transitive, that is the map $f\colon G\times X \to X\times X$ given (on generalized elements) by $f(g, x) = (gx, x)$ is an isomorphism. This is going to be some relatively elementary category theory but I think it’s worth writing it up. First a few facts about isomorphisms in toposes, they can be found for example in Sheaves in Geometry and Logic.¹

Epimorphisms in a topos are stable under pullback.

In a topos every morphism $f\colon X\to Y$ has a functorial factorization $f = m\circ e$ with $m$ a monomorphism and $e$ and epimorphism.

A morphism $f$ is an isomorphism if and only if $f$ is both monic and epic.

Now, let $f\colon A\to B$ be a local monomorphism, i.e. there is an epimorphism $U\to *$ such that the pullback $f\times U$ of $f$ to $U$ is a monomorphism. Then, since epimorphisms are stable under pullback, it follows that in the commutative square $\begin{tikzcd} A\times U \ar[into, r, "f\times U"] \ar[onto, d] & B\times U \ar[onto, d] \\ A \ar[r, "f"'] & B \end{tikzcd}$ both vertical maps are epimorphisms. Now let $\varphi,\psi\colon T\to A$ be a pair of morphisms such that $f\varphi = f\psi$ . Then, denoting by $\varphi_U$ and $\psi_U$ the pullbacks to $U$ , we have $f_U\varphi_U = f_U\psi_U$ . but $f_U$ is a monomorphism by assumption, so $\varphi_U = \psi_U$ . So we have a commutative diagram $\begin{tikzcd} T\times U \ar[r, "\varphi_U = \psi_U"] \ar[onto, d] & A\times U \ar[onto, d] \\ T \ar[r, "\varphi", shift left] \ar[r, "\psi"', shift right] & A \end{tikzcd}$ in which the vertical maps are epimorphisms. It follows that $\varphi = \psi$ .

Now, if $f$ is a local epimorphism, then again we have the diagram $\begin{tikzcd} A\times U \ar[onto, r, "f\times U"] \ar[onto, d] & B\times U \ar[onto, d] \\ A \ar[r, "f"'] & B \end{tikzcd}$ and it is immediate that $f$ is an epimorphism. In summary:

In a topos, any local epimorphism is an epimorphism, any local monomorphism is a monomorphism, and any local isomorphism is an isomorphism.

Now, let’s check that $G$ –torsors $X$ as defined above are free and transitive. Take an epimorphism $U\onto *$ such that $X\times U$ is trivial in $\ca C/U$ . The action of $G$ on itself by left multiplication is plainly free and transitive, so in $\ca C/U$ we have the isomorphism $(G\times U)\times_U (X\times U) \iso (X\times U) \times_U (X\times U)$ and $(G\times U)\times_U(X\times U) = (G\times X)\times U$ and $(X\times U)\times_U (X\times U) = (X\times X)\times U$ because pullback preserves products. So, $G\times X\to X\times X$ is a local isomorphism, hence an isomorphism.

Saunders Mac Lane, Ieke Moerdijk. Sheaves in geometry and logic. Springer, 1994. ISBN: 0-387-97710-4↩