Problem in understanding composite 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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As @Aryabhata pointed out in an earlier question of yours, please avoid posting images in this way. And this image is particularly unreadable. Why can you not just type the question out? After all, it's just 4 lines long. – Srivatsan Sep 11 '11 at 17:09
It is a matter of following the arrows; the convention is that you evaluate first the rightmost expression , and then the one to its left. Then, in (c), you first have f going from A to B , then $f^{-1}$ goes from B to A , so, you take an input 'a' from A, map it to b in B using f, then you have $f^{-1}$ , which goes from B to A, so the result is a function from A to A. – gary Sep 11 '11 at 17:11
@Srivatsan I think OP is not familiar with tex, and so such problems are a difficult to type. Seeing as his questions always get edited to avoid the image, there is little incentive for him to change this habit. – Austin Mohr Sep 11 '11 at 17:43
@Austin I agree there is little incentive for the OP to change his/her (:)) habit, and that's part of the problem. Also, even if not in TeX, s/he could at least type it in plain text, may be with A instead of $A$ like in the last paragraph. This will at least save the editors the trouble of typing out the whole text from the picture. – Srivatsan Sep 11 '11 at 17:48
I do not know how you all right latex. I am sorry but I have no idea how to write it. – Fahad Uddin Sep 11 '11 at 22:35

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] – Mark Bennet Sep 11 '11 at 17:39

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.

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