Inverse and composite functions.

Question

I do not understand the following.

If $f$ is a bijection, the $f \circ f^{-1}$ is the identity function on $Y$and $f^{-1} \circ f $ is the identity function on $X$.

I just thought that the identity function is essentially $y=x$?

What does it mean for the ' identity function to be on $Y$ or $X$?'

Answer 1

They mean that it becomes a function from $X$ to $X$ (for the second identity function), or from $Y$ to $Y$ (for the first identity function).

For example, for the second one: $(f^{-1} \circ f) (x) = x$ for any $x \in X$, hence it maps any $x \in X$ to $x$ itself. Hence it is the identity function from $X$ to $X$, which they say is a function 'on' $X$ itself: any function from $X$ to $X$ itself is a function 'on' $X$, whether it is an identity function or not.

Answer 2

Identity function maps an element to itself, so it is not possible to exist an identity function from X to Y. The sentence "identity function to be on Y or X" means that it depends on your domain. If your domain is X then you should consider the function f and that is why the identity map is from X to X.

Answer 3

Suppose $f: X \to Y$ is a bijection.

Then $f$ has an inverse $f^{-1}: Y \to X$.

Furthermore, $f^{-1} \circ f: X \to X$ is the identity function from $X$ to $X$. To see why this is, first of all, we apply $f$ on an element $x \in X$ which gives $f(x) \in Y$ by definition. Next, $(f^{-1}\circ f)(x) = f^{-1}(f(x)) \in X$, hence $f^{-1} \circ f: X \to X$. Similarly, you may show that $f \circ f^{-1}: Y \to Y$.

Next, by definition of an inverse function, suppose for $f: X \to Y$ that $f(x) = y$ for $x \in X$ and $y \in Y$. Then, $f^{-1}(y) = x$, hence $$(f \circ f^{-1})(y) = f(f^{-1}(y)) = f(x) = y$$ hence $f \circ f^{-1}$ is the "identity" function from $Y \to Y$. The $X \to X$ case is similar.