1. I am a total beginner to category theory, so might abuse common notation or terminology at some point without even being aware of that. So please stay sharp. My primary goal is to master notation, make myself clear and understandable by others.
  2. I know my proof technique is insufficient, I haven't had enough practice; so I'll try to perform a fully correct proof by contradiction and would like to get some feedback whether it looks like a real proof or not. If not, it would be really great to see how it supposes to look.

Show that a map in a category can have at most one inverse.

Given somewhat category $\mathbb{C}$ with $A, B \in \mathsf{Ob}(\mathbb{C}), A \not = B$ and $f \in \mathsf{Hom}_{\mathbb{C}}(A, B)$, there can't be more than a single $g \in \mathsf{Hom}_{\mathbb{C}}(B, A)$ such that $g \circ f = 1_A, f \circ g = 1_B$.

Since I decided to prove the statement above by contradiction, I start from assumption that there indeed exists $g' \in \mathsf{Hom}_{\mathbb{C}}(B, A), g' \not = g$ such that $g' \circ f = 1_A, f \circ g' = 1_B$. Actually, I kind of uncertain here, because there is no explicit way how to calculate equality of morphisms; maybe there is one, however I did not mention it and thus $g' \not = g$ seems to be useless.

Nevertheless: $$g \circ f = 1_A = g' \circ f$$ $$\Rightarrow g \circ f = g' \circ f$$ $$\Rightarrow g = g'$$

Hence, $g = g'$ and $g \not = g'$ is contradiction. Thus, my initial assumption can't be true; thus, $g = g'$; thus, there indeed can't be more than a single inverse of $f$.

  • $\begingroup$ How do you get from $fgf=fg'f$ to $gf=g'f$? $\endgroup$ – Lord Shark the Unknown Feb 25 '18 at 11:52
  • $\begingroup$ @LordSharktheUnknown, looks like I even don't need that. I simplified a bit. Now that step becomes a natural consequence of composition: since $f$ is the same on both right and left sides, I am allowed to ignore it, preserving the equality. $\endgroup$ – Sereja Bogolubov Feb 25 '18 at 12:01
  • $\begingroup$ Now, how do you get from $gf=g'f$ to $g=g'$? $\endgroup$ – Lord Shark the Unknown Feb 25 '18 at 12:07
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    $\begingroup$ See the answer by lush. $\endgroup$ – Lord Shark the Unknown Feb 25 '18 at 12:21
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    $\begingroup$ You are not doing a proof by contradiction but a proof of a negation, see this, though it will take a double negation elimination to go from the statement you're trying to prove to the fact that any two inverses are equal. That extra step would turn your (attempted) proof into a proof by contradiction of the uniqueness of inverses, but your (attempted) proof contains a direct proof of that statement, so that would be silly. Generally, proof by contradiction is not a virtue. $\endgroup$ – Derek Elkins Feb 25 '18 at 19:50

If $g\circ f=g'\circ f$ then the conclusion $g=g'$ that you make is only justified under the extra condition that $f$ is right-cancellable, or - in terms of categories - that $f$ is an epimorphism.

Correct route:

If $g\circ f=1_A$ and $f\circ g=1_B$ and next to that also $g'\circ f=1_A$ then$$g'=g'\circ1_B=g'\circ (f\circ g)=(g'\circ f)\circ g=1_A\circ g=g$$

  • $\begingroup$ Isn't every isomorphism an epimorphism too? And also a monomorphism? $\endgroup$ – jaspreet Feb 27 '18 at 3:28
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    $\begingroup$ @jaspreet Every isomorphism is a split epimorphism and a split monomorphism which is a much stronger claim than simply being an epimorphism or monomorphism. Right-cancellable means split epimorphism not just epimorphism, i.e. $f$ is an epimorphism is a much weaker statement than $f$ is right-cancellable. (For example, the axiom of choice is the statement that all epimorphisms in $\mathbf{Set}$ are split.) Of course, we can't use the fact that isomorphisms are epimorphisms/monomorphisms because the calculation in the answer is part of the proof of that fact! $\endgroup$ – Derek Elkins Mar 1 '18 at 0:04
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    $\begingroup$ @DerekElkins It does not make sense. A morphism $f:A\rightarrow B$ is defined to be an epimorphism if and only if $\forall \ object \ T, \forall t_1, t_2:B\rightarrow T, (t_1\circ f=t_2\circ f)\Rightarrow (t_1=t_2)$. It makes sense to call epimorphism to be a right cancellable morphism, as we are cancelling out fs in the implication. Whereas, split means there exists a section for $f$. Wikipedia seems to agree with this definition: en.wikipedia.org/wiki/Epimorphism Hence, I do not agree with you on "Right cancellabe means split epimorphism not just epimorphism". $\endgroup$ – jaspreet Mar 2 '18 at 20:47
  • $\begingroup$ @jaspreet You're right. I took "right-cancellable" as more like "right invertible". $\endgroup$ – Derek Elkins Mar 2 '18 at 21:34

I don't yet have the reputation to write a comment, so I'll just write that here: You don't need the assumption, that the objects $A$ and $B$ are not equal. It is no problem to have morphisms $A \to A$ in a category.

Now it is not in general true, that $g \circ f = g' \circ f \Rightarrow g = g'$. Therefore you should not use such an argument.

If you want a little hint: Try to start with $g = ...$ and deduce that $g = g'$ just by the information you've got.

//Edit: Not really important, but maybe still worth mentioning: The symbol $\mathbb{C}$ is very non-standard for a category.


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