I try to learn the theory of category from The Joy of Cats. I got stacked with the first exercise (3A a).

If we have a simple graph with one vertex and 2 nodes all we know is that in category there is 2 objects and one homomorphism (beside identities). Hence:

\[C = (\{\{x, y\}, \{1, 2\}\}, \{f, id\}, id, \cdot), f = \{(x, 1), (y, 2)\}\]


\[C = (\{\{x\}, \{1\}, \{f, id\}, id, \cdot), f = \{(x, 1)\}\]

Should have the same graph but are they isomorphic?

Note about notation: I treat category as quadruple $(O, H, id, \cdot)$ where $O$ is class of objects (here is set), $H$ is class of homomorphisms (in example set). $id$ is identity and $\cdot$ is composition of homomorphisms.


It is a bit hard to understand your notation. If understand correctly, the isomorphism is realized by the functor $F$ which sends $\{x, y\}$ to $\{ x \}$ and $\{ 1, 2 \}$ to $\{ 1 \}$ (on objects) and send the unique non trivial morphism $f$ to $f$.

  • $\begingroup$ There should be { brackets, but they are lost in the formatting. On MO one can avoid that by using backticks (`) around formulas, but here it does not work. Any other workaround? $\endgroup$ – Andrea Ferretti Aug 5 '10 at 15:55
  • $\begingroup$ Prefix the $\\{$ by two \\ - then you get a single bracket. $\endgroup$ – Maciej Piechotka Aug 5 '10 at 16:04
  • $\begingroup$ As I understend you answer you mean that isomorphism is when there is 'bi-jective' functor $F$ from one category to another. $\endgroup$ – Maciej Piechotka Aug 5 '10 at 16:07

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