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This question is only about terminology. Inside a category we have the standard wordings:

  1. An arrow $f: X \rightarrow Y$ is an isomorphism if there is another arrow $g: Y \rightarrow X$ such that $g \circ f = id_X$ [added] and $f \circ g = id_Y$.

  2. Two objects $X,Y$ are isomorphic if there is an isomorphism $f: X \rightarrow Y$.

I wonder whether there are comparably catchy names for the corresponding functorial cases. All I could find so far is:

  • A functor $F : C \rightarrow D$ yields an equivalence of categories if there is another functor $G : D \rightarrow C$ such that $G \circ F \simeq \mathsf{I}_C$ and $F \circ G \simeq \mathsf{I}_D$ (plus further conditions)

I already find this terminology rather clumsy, but I did not find at all a phrase that could replace $\dots$ in the following definition:

  • Two objects $X\in C,Y \in D$ are $\dots$ if there is an "equivalence functor" $F: C \rightarrow D$ such that $F(X) = Y$.

Did I just miss something?

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ad 1: "... and $f\circ g=id_Y$" – Hagen von Eitzen Nov 12 '12 at 16:19
To clarify @HagenvonEitzen's comment: Your definition for isomorphism is wrong. – Thomas Andrews Nov 12 '12 at 16:22
Why would you expect such a terminology? If $R$ and $S$ are rings, and $f:R\to S$ is a ring isomorphism, do you have a term for the relationship between $r\in R$ and $f(r)\in S$? In fact, this relationship is only made valid under $f$ - there might be another isomorphism of rings that doesn't send $r$ to the same $f(r)$ – Thomas Andrews Nov 12 '12 at 16:24
@Hagen, Thomas: I added the missing part. – Hans Stricker Nov 12 '12 at 16:28
This kind of relationship never gets a name (because experience shows it doesn't need one) beyond, sometimes, "corresponding objects" leaving implicit under which equivalence they correspond. Here are more examples: what do you call $x$ and $y$ when $f(x)=y$ for $f$ some (a) bijection of sets, (b) isomorphism of groups, (c) diffeomorphism of smooth manifolds, (d) invertible linear transformation? Really, it's fine to say "corresponding under $f$" for all of these cases. – Omar Antolín-Camarena Nov 13 '12 at 15:23

Well, it is not usual to relate two objects from different categories. I would say, that $Y$ is the 'correspondent' of $X$, or something like this..

However, one can still say that $X$ and $Y$ are isomorphic, due to the following fact:

Categories $C$ and $D$ are equivalent if and only if there is a category $E$ and full embeddings $\phi:C\to E$ and $\psi:D\to E$ such that for all $X\in C$ there is an $Y\in D$ 'isomorphic to $X$ in $E$', meaning that $\phi(X)\cong \psi(Y)$ in $E$, and for all $Y\in D$ there is an $X\in C$ isomorphic to $Y$ in $E$.

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+1 Nice observation. – amWhy Nov 12 '12 at 17:51
What is the category $E$ you have in mind? The most natural one I can think of is the comma category $(F \downarrow \mathcal{D})$, but it lacks symmetry... – Zhen Lin Nov 12 '12 at 18:19
@Berci: In any case, $X$ and $F(X)$ are related by the equivalence $F$, aren't they? I try to understand why this relation is so "unimportant" per se that it doesn't deserve a name. – Hans Stricker Nov 12 '12 at 20:11
@Hans: Well, if you have an isomorphism between groups, how do you call the correspondent elements? I think, they are called 'correspondents'... – Berci Nov 13 '12 at 14:53
@Zhen: Consider an adjoint equivalence of functors $F:A\to B$ and $G:B\to A$ with isomorphisms $\eta:1_A\to FG$ (unit) and $\varepsilon:GF\to 1_B$ (counit), then, using these, one can glue together the collage of profunctors $F_*$ and $G_*$. Another way is to select skeletons in both A and B, and an isomorphism between them, and 'connect' the corresponding objects by invertible new pair of arrows for each correspondence. – Berci Nov 13 '12 at 14:57

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