What are morphisms in the category of sets $\mathbf{Set}$? Do i understand correctly that morphisms in the category of sets $\mathbf{Set}$ are ordered triples $(f, A, B)$ where $f$ is a function $A\to B$?
It seems that it is often claimed, even in the Categories for the working mathematician by Mac Lane, that morphisms are functions, while this is obviously wrong, as the codomain operation on morphisms would not be possible to define (corectly) in this case.
(Incidentally, a similar error reappears in the definition of the category of functors: natural transformations are taken to be morphisms, but probably the correct definition of morphisms would be all triples $(\tau, S, T)$, where $\tau$ is a natural transformation from the functor $S$ to the functor $T$.)
 A: Alexey you're observation is correct. To or to be exact in MacLane's book a function $f \colon X \to Y$ is thought as an ordered triple of the form $f=\langle X,Y,\bar f\rangle$ where $X$ and $Y$ are sets and $\bar f \subset X \times Y$ is a functional relation (i.e. a relation such that for every $x \in X$ there's a unique $y \in Y$ with the property $(x,y) \in \bar f$) [as written in the introduction of the book].
In such case the operation of codomain is defined, being just the operation giving the second element of the ordered triple.
Clearly the same objection applies to the case of natural transformations and functors, but that's just to avoid to be too pedantic.
A: To be precise, an ordinary function itself is always a triple. The domain and the codomain of a function is always part of its definition.
Two functions with the same expression, and isomorphic domain or codomain, can only be "completed" to a similar function in 2 different ways.  In general, all the "static parts" of X are always part of the definition of X.
It's not a trivial idea, I think this completion appears in non trivial context.
