What conditions hold $f\left(f^{-1}(x)\right)=x$

Question

For example

$$\begin{align*}\text{sin(arcsin(x))=x}\end{align*}$$

I think this is a trival question maybe asked too many times, where are they?

score 3 · Answer 1 · answered Jul 15 '13 at 06:24

3

$$f(f^{-1}(x)) = x$$ is always true provided $f^{-1}(x)$ exists in the first place, i.e., there exists at least one element $y$ such that $f(y) = x$.

answered Jul 15 '13 at 06:24

2

(And even then it can be more generally interpreted with images and preimages.) – anon Jul 15 '13 at 06:26
@anon What if $x$ does not belong to the image of $f$? – Jul 15 '13 at 06:27
@user17762 The domain of $f\circ{f^{-1}}$ is by definition the image of $f$. – rurouniwallace Jul 15 '13 at 06:29
@AWertheim That sounds more or less identical to what I said in my comment. – rurouniwallace Jul 15 '13 at 06:36
@ZettaSuro, my apologies. I had thought your comment was a correction to user17762's answer, not an affirmation of it. – Alex Wertheim Jul 15 '13 at 06:42
@AWertheim It's my fault, I worded it poorly. I should have said "Yes, $f\circ{f^{-1}}$ must be the image of $f$". – rurouniwallace Jul 15 '13 at 06:45

score 2 · Answer 2 · answered Jul 15 '13 at 06:37

In my opinion, it doesn't make sense to talk about $f^{-1}$ of an element unless we know that $f$ is bijective. Usually, it makes more sense to talk about pre-image of a set. For example, you might mean $f^{-1}(\{x\})$ instead of just $f^{-1}(x)$.

Therefore, it is always true by definition that $f(f^{-1}(A))\subseteq A$ with equality if $f$ is surjective. In the same spirit, $f^{-1}(f(B))\supseteq B$ with equality if $f$ is injective.

The problem arises in your case when $f$ need not be surjective. Then in this case the pre-image $f^{-1}(\{x\})$ may be empty. For example, if $f:\mathbb{R}\rightarrow\mathbb{R}$ with $f(x)=x^{2}$, then $f^{-1}(\{-1\})=\emptyset$. Therefore, $f(f^{-1}(\{-1\}))=\emptyset\subseteq\{x\}$. Thinking of pre-image of a set rather than an element prevents confusion.

score 1 · Answer 3 · answered Jul 15 '13 at 06:30

This is a definition of a (left) inverse of a map. It is less confusing to look at this when domain a co-domain are different, or at least denoted by different symbols. That is, let us consider a map $f:X\to Y$, then a map $g:Y\to X$ is called left (right) inverse, if $g(f(x)) = x$ for any $x\in X$ (if $f(g(y)) = y$ for all $y\in Y$).

Both inverse functions are often denote by the same symbol $g = f^{-1}$, so that the equation $f(f^{-1}(x)) = x$ that you wrote in fact holds for any $x$ in the co-domain of $f$ (that is, in the set $Y$ in the previous notation). For example, even though $f = \sin$ seem to send reals to reals, its range is only $[-1,1]$ and that's where the $\arcsin$ is defined (so that although $X = \Bbb R$, here $Y = [-1,1]$). In particular, $\sin$ does not allow for the right inverse in case $Y'=\Bbb R$.

What conditions hold $f\left(f^{-1}(x)\right)=x$

3 Answers3