Abelian category equivalent to a non-abelian category I was told that if we have an equivalence of categories $F : \mathcal{A} \rightarrow \mathcal{B}$ with $\mathcal{A}$ abelian, then it is not necessarily true that $\mathcal{B}$ is also abelian.
I would like to know if there are nice examples of an abelian category $\mathcal{A}$ which is equivalent to a non-abelian category $\mathcal{B}$.
Furthermore, are there any conditions over $F$ or $\mathcal{B}$ so that we have "$F : \mathcal{A} \rightarrow \mathcal{B}$ is an equivalence and $\mathcal{A}$ is abelian implies $\mathcal{B}$ is abelian"?
 A: Here is a manifestly invariant definition of an abelian category $\mathcal{C}$. It is a category with finite limits and colimits such that:


*

*(It is pointed) the map from the initial to the final object is an isomorphism; we denote by 0 any object which is both initial and final.

*(It is semiadditive) the map $X \amalg Y \to X\times Y$, given on $X$ by components $(\mathrm{id}_X, X \to 0 \to Y)$ and on $Y$ by components $(Y \to 0\to X, \mathrm{id}_Y)$, is an equivalence. We denote by $X\oplus Y$ the coproduct or product, identified as above. This equivalence produces an abelian monoid structure on all hom sets, where addition arises from $X \to X\oplus X \stackrel{f\times g}{\to} Y\oplus Y \to Y$.

*(It is additive) the shearing map $X\oplus X \to X\oplus X$, given by adding the identity map to the projection onto the first component followed by inclusion, is an equivalence. Equivalently, each hom-monoid has the property that it is group-like.

*(first isomorphism theorem) if $f: A \to B$ is arbitrary, then the map $A/\mathrm{ker}(f) \to \mathrm{ker}(B \to B/A)$ is an isomorphism.


Being an abelian category is a property not structure.
A: What you were told is wrong, for we have the following:
Proposition. If two categories are equivalent and one of them is abelian, then so is the other.
A proof (and some related results) can be found in Satz 16.2.4 in H. Schubert, Kategorien II, Springer, 1970 (likewise in the English version https://www.amazon.com/Categories-Horst-Schubert/dp/3642653669, under the same numbering). 
A: As a discerning dissenting voice, let me say that it might be true, depending on your definitions.
Take abelian $A$. Let $B:=A$ with forgotten enrichment in abelian groups. Then the identity functor is equivalence, but $B$ is not abelian because it is not even additive.
There are two definitions in play. The first is abelian category. The standard (ncatlab or Wikipedia) definition of abelian category asks for the category to be pre-additive, which requires enrichment in abelian groups that can be forgotten. An alternative definition, mentioned in other answers, is from Freyd's book: there abelian is an inherent property rather than an additional structure. This definition would make the statement false.
The second definition is  non-abelian category. There is no accepted definition. One possibility is to think of this as just a category. Together with the first definition of abelian category, this makes the statement true. An alternative is to think of a category that cannot be abelian (essentially the negative of the second definition). This choice would make the statement false.
