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Reading through the Category Theory Wikibook, I came across the following exercise:

(Harder.) If we add another morphism to the above example, it fails to be a category. Why? Hint: think about associativity of the composition operation. Two objects A and B, with an arrow going from A to A (labeled idA), an arrow from A to B labeled g, an arrow from B to B labeled idB, and two arrows from B to A, one labeled f and the other labeled h.

I haven't been able to figure out why the diagram given is not a category. The circles are the objects, the arrows are the morphisms, and there is a clear notion of composition.

Going through the category laws (in reverse order):

1. There is an identity element for every object ($id_A$ and $id_B$).

2. The set of morphisms are closed under composition: $f \circ g > = id_A$, $h \circ g = id_A$, $g \circ f = id_B$, $g \circ h = id_B$, and any composition involving an identity morphism is clearly (I think) closed. Compositions like $f \circ h$ are meaningless.

3. Coming to the hint in the prompt, and trying to prove associativity, $h \circ (g \circ f) = g \circ id_B = g$, whereas $(h > \circ g) \circ f = id_B \circ f = f$, so $h \circ (g \circ f) \neq (h > \circ g) \circ f$.

This is where I get a little confused, though, since $f$ and $h$ are in some ways "the same", since A and B are isomorphic (right?). I haven't figured out exactly why I'm confused about this, but it seems like this exercise means that there can only ever be a single morphism between two objects in a category, which is clearly nonsense. What am I missing in all this?

Thanks!

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    $\begingroup$ I think the exercise might mean that if $h$ is different from $f$, more morphisms will be needed to make it a category. $\endgroup$
    – Tunococ
    Aug 13, 2014 at 12:29

2 Answers 2

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$f$ and $h$ are distinct simply because the diagram says so. Any argument a la "They must be the same beacuase they are the unique twosided inverse of the isomorphism $g$" already assumes that we have a category.

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  • $\begingroup$ OK, that makes sense. Thanks. $\endgroup$
    – anjruu
    Aug 13, 2014 at 12:42
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Well the statement of the problem is not extremely clear, perhaps on purpose, but what it is trying to teach is the following: in the "example above" you conclude - by looking at the diagram without the $h$- that the composition of $f$ and $g$ gives an identity (on $A$ and on $B$). If they now tell you to add another - different - morphism $h$, you must be sure that the compositions $g \circ h$ and $h \circ g$ exists. Since the only appropriate and available morphisms are the identities, you are obliged to write: $g \circ h=id_B$, but then $$g \circ h=id_B=g \circ f$$ and , by left-composing with $f$ you get the contradiction $f=h$. Indeed $$f\circ g \circ h=id_A \circ h=h=f \circ id_B=f=f \circ g \circ f$$ So you cannot add a different morphism $h$ to the category in the "example above". Put it in another way: you cannot have two different 2-sided inverses ($f$ and $h$) to a morphism $g$.

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  • $\begingroup$ OK, the statement about the different two-sided inverses explains a lot, thanks. $\endgroup$
    – anjruu
    Aug 13, 2014 at 13:08

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