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Let $F=(f;A;B)$ is a morphism of the category $\mathbf{Rel}$ (the category whose objects are sets and morphisms are defined as binary relations).

How to name and how to denote $f$ when we know $F$?

I propose to call $f$ the graph of $F$. Right name?

But how to denote it? Are there a standard notation?

I propose the following (non-standard) notation: $\mathrm{GR}\, F= f$.

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    In practice you won't have to distinguish between $f$ and $F$. And I would call $f$ just a relation from $A$ to $B$.2012-05-27

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Graph of F is perfectly good.

I would write: $Graph(F)=f$, since GR in itself is a bit opaque as a name.

Edited after taking comments into account:

$\mathbf{Rel}$ is a bit atypical in the sense that it is named after its morphisms, while most categories are named after their objects (see MacLane's CWM chapter 1 notes). We can however form the category of arrows of $\mathbf{Rel}$, ie. $\mathbf{Rel}^\mathbf{2}$ (see CWM again) where the relations are the objects. So we could then write:

$U(F)=f$

Where $U:\mathbf{Rel}^\mathbf{2}\to \mathbf{Set}$ is the forgetful functor from $\mathbf{Rel}^\mathbf{2}$ to $\mathbf{Set}$ (since the graph of F is a set).

$\mathbf{2}$ is the category with 2 objects and just one morphism between them

Correction: On second thought: graph of F is not a good name. That is because graph is normally meant to be an ordered pair (Wikipedia), while your $f$ is just a subset of $A \times B$. So I would simply call $f$ "the underlying set of $F$" and use the $U(F)=f$ notation to derive it

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    I'm tempted to apply a -1: the category $\textbf{Rel}$ is _not_ a category whose objects are relations.2012-05-28
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    There is no functor U here ...2012-05-28
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    @ZhenLin Ops...you are very right. My answer was not very well thought out. I would prefer to delete it if possible or otherwise see if there is a possible category where relations are objects and morphisms are...what?2012-05-28
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    There is a _2-dimensional_ category $\mathfrak{Rel}$, and the hom-category $\mathfrak{Rel}(X, Y)$ is an ordinary category whose objects are relations from $X$ to $Y$ and morphisms are, well, maps of sets. But this is just a really fancy way of talking about $\mathscr{P}(X \times Y)$.2012-05-28
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    @zhenLin yes or we could consider this: $\mathbf{Rel'}$: objects are relations (ordered triples), morphisms from to are with $r' \circ f =r \circ f'$ where composition in $\mathbf{Rel}$ is meant. This is called the category of arrows of $\mathbf{Rel}$ (MacLane's CWM page 40)2012-05-28
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    What you have defined is not a functor. The morphisms of $\textbf{Rel}^\mathbf{2}$ are still relations and need to go somewhere in $\textbf{Set}$.2012-05-28
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    @ZhenLin indeed I only wrote the object part of an hypothetical functor. The morphism part is not too obvious (at least to me). Maybe it does not even exist. Can someone prove it one way or the other?2012-05-29