identity function

Sep 2009
502
39
I found the following from a book with no proof:

Let \(\displaystyle f:A \rightarrow A\) be defined by the formula \(\displaystyle f(x)=x\), then \(\displaystyle f\) is called the identity function, denoted by \(\displaystyle 1\) or by \(\displaystyle 1_A\).

Let \(\displaystyle f:A \rightarrow B\) and it has the inverse function \(\displaystyle f^{-1}:B\rightarrow A\), then \(\displaystyle f ^{-1}\circ f=1\)

Question: Is it true that \(\displaystyle f ^{-1}\circ f=1\)?

Isn't \(\displaystyle f(x)^{-1} \circ f(x)=x\)?
 
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Plato

MHF Helper
Aug 2006
22,455
8,631
I found the following from a book with no proof:

Let \(\displaystyle f:A \rightarrow A\) be defined by the formula \(\displaystyle f(x)=x\), then \(\displaystyle f\) is called the identity function, denoted by \(\displaystyle 1\) or by \(\displaystyle 1_A\).

Let \(\displaystyle f:A \rightarrow B\) and it has the inverse function \(\displaystyle f^{-1}:B\rightarrow A\), then \(\displaystyle f \circ f^{-1}=1\)

Question: Is it true that \(\displaystyle f \circ f^{-1}=1\)?

Isn't \(\displaystyle f(x) \circ f^{-1}(x)=x\)?
Are you sure that you have not turned things around here.
For one, if \(\displaystyle f:A\to B\) then \(\displaystyle f \circ f^{-1}:B\to B\).
So what are you asking?
 
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Sep 2009
502
39
Are you sure that you have not turned things around here.
For one, if \(\displaystyle f:A\to B\) then \(\displaystyle f \circ f^{-1}:B\to B\).
So what are you asking?
Yes, made a mistake. Sorry.

Question: Is it true that \(\displaystyle f^{-1}\circ f =1\)?
 

Plato

MHF Helper
Aug 2006
22,455
8,631
But it is just a matter of notation: \(\displaystyle f\circ f^{-1}=1_B~\&~ f^{-1}\circ f=1_A \).
Recall that \(\displaystyle \left( {\forall x \in A} \right)\left[ {1_A (x) = x} \right]\) thus \(\displaystyle f^{-1}\circ f(x)=1_A(x)=x\).
 
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Sep 2009
502
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But it is just a matter of notation: \(\displaystyle f\circ f^{-1}=1_B~\&~ f^{-1}\circ f=1_A \).
Recall that \(\displaystyle \left( {\forall x \in A} \right)\left[ {1_A (x) = x} \right]\) thus \(\displaystyle f^{-1}\circ f(x)=1_A(x)=x\).
I found \(\displaystyle i_A\) a much better notation. Perhaps, the book is too old. The inverse function denoted by 1 is terribly confusing. I have mistaken it as a number.