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As an example, let me draw a picture with two objects $A$ and $B$, and two arrows $$A \xrightarrow{\;\;f\;\;} B \quad\text{and}\quad A \xleftarrow{\;\;g\;\;} B\,.$$ I see two natural ways to interpret this drawing:

  1. It could be the category with two objects that contains exactly four morphisms $f$,$g$,$1_A$, and $1_B$, where it's implied that $fg = 1_B$ and $gf=1_A$.

  2. Or it could be the category "freely generated" by this picture, where there are no relations among the morphisms besides those necessary to satisfy the axioms. This way, you get all the possible composite morphisms like $fgfgfgfg$ and the like.

In my experience the second interpretation is the usual one. But is there any commonly used phrasing to specify this second interpretation? I just want to stay consistent with any precedent there might be. Saying "the category freely generated by ..." or "the category with no relations" don't feel correct, like those should each refer to some other idea.

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    $\begingroup$ "the category freely generated by these arrows" is perfectly correct and common. $\endgroup$ Commented Jan 22, 2018 at 3:35
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    $\begingroup$ There is even an adjunction from the category of graphs to the category of categories mapping a graph to the corresponding category freely generated by it, etc $\endgroup$ Commented Jan 22, 2018 at 3:36
  • $\begingroup$ @MarianoSuárez-Álvarez Thank you! It looks like I was a big too worried about using the word "free" here. $\endgroup$ Commented Jan 22, 2018 at 3:55

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Contrary to your feeling, "the category freely generated by..." is exactly the normal way to say this. To be completely precise, you could say that it is the free category on the quiver with two vertices $A$ and $B$ and two arrows $f:A\to B$ and $g:B\to A$. (There is a forgetful functor from categories to quivers, and this notion of "free category" is its left adjoint.)

Another phrase sometimes used is "the walking ...". For instance, the category freely generated by an object $A$ and an endomorphism $f:A\to A$ (i.e., the monoid category $\mathbb{N}$) could be called "the walking endomorphism", since a functor from this category to a category $D$ is the same thing as an endomorphism of some object of $D$. In your particular example, though, I don't see a nice way to use this phrasing, since there's not really a word for a pair of maps going in opposite directions.

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    $\begingroup$ For the first scenario, though, you could talk about the "walking isomorphism". $\endgroup$ Commented Jan 22, 2018 at 4:02

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