Are (quasi)topoi Barr-exact? Question is in the title. I know topoi are extensive (https://ncatlab.org/nlab/show/toposes+are+extensive) and I believe they are exact too but I have failed to see how it follows from the definition or to find a reference.
Also, it seems to me that quasitopoi have most of the good properties of topoi and so should be exact too, or at least regular.
I am not restricting myself to Grothendieck (quasi)topoi.
 A: See Chapter 1, Section 23: "Exactness Properties of Quasitopoi" of Oswald Wyler's Lecture Notes on Topoi and Quasitopoi.
There it is proven that 


*

*Pullbacks in a quasitopos preserve epimorphisms and strong epimorphisms

*Every morphism in a quasitopos can be factored as an epimorphism followed by a regular monomorphism, and as a regular epimorphism followed by a monomorphism.


In particular, quasitopoi are regular. 
On the other hand, a pair of morphisms $R\rightrightarrows X$ is the kernel pair of $X\xrightarrow{f}Y$ if and only if the fork $R\to X\times X\underset f{\overset f\rightrightarrows}Y$ is an equalizer (this is the general construction of pullbacks via equalizers to a product). In particular, effective equivalence relations (ones that are kernel pairs) are always regular in the sense that the induced $R\hookrightarrow X\times X$ is a regular monomorphism. What is true in a quasitopos (see Section 15) is the converse: every regular equivalence relation is the kernel pair of its characteristic morphism $X\to PX$, so regular and effective equivalence relations coincide in quasitopoi.
Thus a quasitopos will fail to be Barr exact precisely when there exists a non-regular equivalence relation, i.e. an equivalence relation that is not represented by the power object. Since in a topos every monomorphism is regular ($U\hookrightarrow X$ is the equalizer of the characteristic morphism $X\to\Omega$ and the composite $X\to\mathbf 1\xrightarrow{\top}\Omega$), it follows that topoi are Barr exact.
