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I was reading a theorem about functions:

Let $f:A\to B$ be any function. Then
$\hskip0.3in$(a) $1_B\circ f=f$.
$\hskip0.3in$(b) $f\circ 1_A=f$.
If $f$ is a one-to-one correspondence between $A$ and $B$, then
$\hskip0.3in$(c) $f^{-1}\circ f=1_A$
$\hskip0.3in$(d) $f\circ f^{-1}=1_B$

Now I am unable to decide that either the input would be from A or B in the part c and d. Previously I used to think that the function mentioned at the right always takes input from A and the one mentioned at the left of composition takes input B but it is not proved from part d of the theorem. Can anyone please suggest how do we get to know what is the input?

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    @Austin: Thanks for the suggestion.2011-09-12

2 Answers 2

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The notation $f: A \rightarrow B$ tells you that the domain (set inputs) is $A$ and the range (set of possible outputs) is $B$. The function goes "from $A$ to $B$", as indicated by the arrow.

An inverse function, such as $f^{-1}$, must go in the opposite direction as the original. In this example, that means $f^{-1}$ goes "from $B$ to $A$". You might also indicate this by writing $f^{-1}: B \rightarrow A$.

The subscripts on the $1$'s indicate their domain. So, $1_A$ is the identity on $A$ and $1_B$ is the indetity on $B$.

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    It might also be worth noting, for the avoidance of doubt, that for the identity the "input" and "output" are the same. So the Identity on A has input and output both equal to A. [Note that it is quite possible that there may be functions from A to A which are not the identity - dependent on what A is - so the identity is a special function, the key properties of which are given in the original post]2011-09-11
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In (c), the input is from A

In (d), the input is from B since $f^{-1} : B \rightarrow A$.

In general, $f : A \rightarrow B$ and $g : B \rightarrow C$. $g \circ f$ takes input from the domain of $f$ (that is $A$) and outputs something from the range (or codomain) of $g$.

Function composition is applied from the left. $f$ is applied first and then $g$. Since $f$ is applied first, the input must come from the domain of $f$ to be well defined.