Equivalence of categories

Remarks

If moreover $G \circ F = 1_{\mathcal{C}}$ and $F \circ G = 1_{\mathcal{D}}$, then $F$ is an isomorphism of categories, a much stronger condition. Example: let $K$ be a field and let $\mathbf{Mat}_K$ have the natural numbers as objects, the $n \times m$ matrices over $K$ as morphisms $m \to n$, and matrix multiplication as composition. The functor $\mathbf{Mat}_K \to$ (finite-dimensional vector spaces over $K$) sending $n$ to $K^n$ and a matrix to its linear map is an equivalence. A quasi-inverse chooses a basis of each space $V$ and sends $V$ to $\dim V$ and a linear map to its matrix in the chosen bases. Non-example: let $\mathbf{1}$ have one object and one morphism, and let $\mathbf{2}$ have two objects and only identity morphisms. No functor $F \colon \mathbf{1} \to \mathbf{2}$ is an equivalence. Both objects $X$ of $\mathbf{2}$ would need an isomorphism $F(G(X)) \cong X$. But $F(G(X))$ is the single object in the image of $F$, and the two objects of $\mathbf{2}$ are not isomorphic. Equivalence of categories is reflexive, symmetric and transitive.