Why Are Inverses Unique in a Category?

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Discussion Overview

The discussion focuses on the uniqueness of inverses in category theory, specifically referencing a section from Steve Awodey's book on isomorphisms. Participants seek to understand and rigorously prove why inverses in a category are unique.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Peter requests a rigorous proof of the uniqueness of inverses in a category, referencing Definition 1.3 from Awodey's book.
  • One participant outlines a proof strategy involving two arrows $g$ and $h$ that serve as inverses of an isomorphism $f$, aiming to show that $g = h$.
  • The same participant elaborates on the proof by considering the composition $hfg$ and demonstrating that it can be expressed in two ways leading to the conclusion that $g = h$.
  • Peter expresses confidence in the correctness of the proof provided and acknowledges the assistance received.
  • Another participant confirms the correctness of Peter's proof and makes a reference to similar reasoning used in Group Theory.

Areas of Agreement / Disagreement

Participants generally agree on the proof strategy and its correctness, but the discussion remains focused on the details of the proof rather than any broader implications or alternative perspectives.

Contextual Notes

The discussion does not address potential limitations or assumptions underlying the proof, nor does it explore alternative definitions or contexts in which inverses may not be unique.

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I am reading Steve Awodey's book: Category Theory (Second Edition) and am focused on Section 1.5 Isomorphisms ...

I need some further help in order to fully understand some further aspects of Definition 1.3, Page 12, including some remarks Awodey makes after the text of the definition ... ...

The start of Section 1.5, including Definition 1.3 ... reads as follows:View attachment 8354In the text of Definition 1.3 we read the following:

" ... ... Since inverses are unique (proof!), we write $$g = f^{-1}$$. ... ... "Can someone please demonstrate a rigorous proof that in a category, inverses are unique ... ?Help will be appreciated ...

Peter
 
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You have a isomorphism $f:A \rightarrow B$ in a category $C$

Suppose, there are two arrows $g,h:B \rightarrow A$ such that

$fg=1_B$, $gf=1_A$, $fh=1_B$, and $hf=1_A$

in other word, each is an inverse of $f$, we have to prove that $g=h$

Consider $hfg:B \rightarrow A \rightarrow B \rightarrow A$

On one hand, we have $hfg=h(fg)=h1_B=h$

On the other hand, we have $hfg= \cdots$, can you finish this ?
 
Last edited:
steenis said:
You have a isomorphism $f:A \rightarrow B$ in a category $C$

Suppose, there are two arrows $g,h:B \rightarrow A$ such that

$fg=1_B$, $gf=1_A$, $fh=1_B$, and $hf=1_A$

in other word, each is an inverse of $f$, we have to prove that $g=h$

Consider $hfg:B \rightarrow A\rightarrow B$

On one hand, we have $hfg=h(fg)=h1_B=h$

On the other hand, we have $hfg= \cdots$, can you finish this ?
Thanks Steenis ...

Hmm ... easy when you see how ... :) ...

We have ...

$$hfg = h(fg) = h 1_B = h$$ ... ... ... ... ... (1)

and

$$hfg = (hf)g = 1_A g = g$$ ... ... ... ... ... (2) ... so it follows that ...

(1) (2) $$\Longrightarrow g = h$$Hope that is correct ...Thanks again Steenis ...

Peter
 
Yes that is correct

You did these things before, for instance, in Group Theory

In post #2, I meant $hfg:B \rightarrow A \rightarrow B \rightarrow A$
Already corrected
 
Last edited:

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