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Apr 15, 2020 at 18:50 comment added Tomáš Jakl The second example that Andrej wrote arises from the adjunction between $\prod$ and $\Delta$ in ncatlab.org/nlab/show/category+of+simple+graphs
Apr 28, 2013 at 13:53 comment added Andrej Bauer I think this proves I am very smart. Had I been doing things randomly, at least some of these would be actual examples. It takes a special talent to produce four plausibly-sounding non-examples.
Apr 28, 2013 at 3:25 comment added Peter LeFanu Lumsdaine In the currently last example, sending $G$ to the discrete graph on $\pi_0(G)$, isn’t there a problem with the unit morphism — edges within each connected component have no edges to map to?
Apr 19, 2013 at 21:49 history edited Andrej Bauer CC BY-SA 3.0
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Apr 19, 2013 at 6:43 comment added Gejza Jenča Please, do not delete the answer. The last example is correct,I think.
Apr 19, 2013 at 6:08 comment added Gejza Jenča The third example is not a functor, I think: take the path with 3 vertices $P_3$. It can be 2-colored, so there is a morphism into $K_2$. But there is no morphism $K_3\to K_2$.
Apr 18, 2013 at 22:12 comment added Andrej Bauer But surely, there are billions and billions of them.
Apr 18, 2013 at 22:11 comment added Andrej Bauer Ugh, maybe I should just delete the answer.
Apr 18, 2013 at 13:34 comment added Todd Trimble I'm not sure the complete graph example works. If you apply such a completion functor to the function $2 \to 1$ from the 2-element set to the 1-element set (with the discrete graph structures), where would the edge in the completion of $2$ go to?
Apr 18, 2013 at 13:26 history edited Andrej Bauer CC BY-SA 3.0
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Apr 18, 2013 at 13:25 comment added Andrej Bauer Oops, that would be a comonad then. Thanks!
Apr 18, 2013 at 12:41 comment added Andreas Blass In the second example, the discrete graph on the same vertices, what is the unit of the monad?
Apr 18, 2013 at 12:34 history answered Andrej Bauer CC BY-SA 3.0