1. ## identity function

I found the following from a book with no proof:

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

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

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

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

2. Originally Posted by novice
I found the following from a book with no proof:

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

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

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

Isn't $f(x) \circ f^{-1}(x)=x$?
Are you sure that you have not turned things around here.
For one, if $f:A\to B$ then $f \circ f^{-1}:B\to B$.

3. Originally Posted by Plato
Are you sure that you have not turned things around here.
For one, if $f:A\to B$ then $f \circ f^{-1}:B\to B$.
Question: Is it true that $f^{-1}\circ f =1$?
4. But it is just a matter of notation: $f\circ f^{-1}=1_B~\&~ f^{-1}\circ f=1_A$.
Recall that $\left( {\forall x \in A} \right)\left[ {1_A (x) = x} \right]$ thus $f^{-1}\circ f(x)=1_A(x)=x$.
But it is just a matter of notation: $f\circ f^{-1}=1_B~\&~ f^{-1}\circ f=1_A$.
Recall that $\left( {\forall x \in A} \right)\left[ {1_A (x) = x} \right]$ thus $f^{-1}\circ f(x)=1_A(x)=x$.
I found $i_A$ 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.