MHB Why Are Inverses Unique in a Category?

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In the discussion on the uniqueness of inverses in category theory, participants seek clarification on a proof from Steve Awodey's book. The focus is on demonstrating that if two arrows, g and h, are both inverses of an isomorphism f, then g must equal h. The proof involves showing that applying the composition of arrows leads to the conclusion that both expressions for hfg yield the same result, thus proving g = h. Participants confirm the correctness of the proof and relate it to similar concepts in Group Theory. The discussion emphasizes the foundational aspect of isomorphisms in category theory.
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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
 
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