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## Comments

\(G_1\): \(f = g\)

\(G_2\): None (I think)

\(G_3\): \(f.h = g.i\)

\(G_4\): None

`\\(G_1\\): \\(f = g\\) \\(G_2\\): None (I think) \\(G_3\\): \\(f.h = g.i\\) \\(G_4\\): None`

For \(\mathcal{G}_2\), shouldn't we impose \(f\) to be the identity morphism, that is \(f = 1\)? Otherwise, there are infinitely many morphisms from \(\bullet\) to \(\bullet\): \(1, f, f \circ f, f \circ f \circ f\), etc.

`For \\(\mathcal{G}_2\\), shouldn't we impose \\(f\\) to be the identity morphism, that is \\(f = 1\\)? Otherwise, there are infinitely many morphisms from \\(\bullet\\) to \\(\bullet\\): \\(1, f, f \circ f, f \circ f \circ f\\), etc.`