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Problem : If $g$ is the inverse function of $f$, then prove that $f(g(x)) = x$.

Solution :

I know it is well known result.

But I have no idea how to prove it.

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rst
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4 Answers4

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I think we all need some enlightenment on what it means to have an inverse for functions.

Let $f : X \to Y$ be a function defined by some expression like $x \mapsto f(x)$ (for instance, when $X=Y= \mathbb R$, we could let $x \mapsto f(x) = 2x+1$).

This function $f$ is said to have a left inverse if there exists a function $g : Y \to X$ such that for every $x \in X$, $g(f(x)) = x$. Similarly, $f$ has a right inverse if there exists $g : Y \to X$ such that for all $y \in Y$, $f(g(y)) = y$.

If we take this definition as the inverse, then if $f$ has a left inverse $g$, if $f(x) = y$, we have $x = g(f(x)) = g(y)$. If $f$ has a right inverse, then if $g(y) = x$, we have $f(g(y)) = f(x) = y$.

Therefore the fact that $f(x)=y \quad \Longleftrightarrow \quad g(y) = x$ is true if and only if $g$ is a left and right inverse for $f$.

If $g$ is simultaneously a left AND right inverse, we can say that $g$ is the inverse of $f$ and denote it by $f^{-1}$ because $f^{-1}(f(x)) = x$ for all $x \in X$ and $f(f^{-1}(y)) = y$ for every $y \in Y$. Let me give some examples to be clearer.

Consider $f : \mathbb R \to [0,\infty[$ defined by $f(x) = x^2$. If $g : [0,\infty[ \to \mathbb R$ is defined by $g(y) = \sqrt y$, then $g$ is a right inverse for $f$, because $f(g(y)) = f(\sqrt{y}) = \sqrt{y}^2 = y$ (because we assume $y \ge 0$ since this is the domain of $g$). Notice that $g$ is not a left inverse for $f$ because $g(f(x)) = \sqrt{x^2} = |x| \neq x$ if $x$ is negative in the domain of $f$.

However, if we look at the function $f : [0,\infty[ \to [0,\infty[$ defined by $f(x) = x^2$ (note that the definition of the map is the same, but the domain is different!), then the function $g : [0,\infty[ \to [0, \infty[$ defined by $g(y) = \sqrt y$ is a left and right inverse for $f$, because then for every positive $x$, $\sqrt{x^2} = x$ and for every positive $y$, $(\sqrt y)^2 = y$.

Feel free to ask any more questions.

Hope that helps,

Patrick Da Silva
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Functions $f(x)$ and $g(x)$ are inverses of one another if: $f(g(x)) = x$ and $g(f(x)) = x,$ for all values of $x$ in their respective domains.

The above property is what inverse functions are defined to be. I don't think a proof is needed for the definition.

Sujaan Kunalan
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  • g is inverse function of $f$ then $f(x)=y$ and $g(y)=x$ then $g(f(x))=g(y)=x$ – rst Jun 18 '13 at 02:08
  • If $g$ is the inverse of $f$, and $f(g(x))=x$, it implies that $f$ is also the inverse of $g$ – Sujaan Kunalan Jun 18 '13 at 02:10
  • @Sujaan : A very trivial proof is required but the ideas are not the same... (if you think in categorical terms this is actually "a statement" that functions are inverses if and only if they are inverses on every element of their domains, even though it is an easy one). – Patrick Da Silva Jun 18 '13 at 04:09
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If $g$ is inverse function of $f$

Then $f(x)=y$ and $g(y)=x$

So $g(f(x))=g(y)=x$

Another Way

If $g$ is inverse function of $f$

Then $f(x)=y$ and $g(y)=x$

So $f(g(y))=f(x)=y$

or $f(g(x))=x$

rst
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    I think you don't understand what a proof means, to be quite honest... I really don't want to be mean, but you sound confused to be. – Patrick Da Silva Jun 18 '13 at 02:42
  • I think proof means to get result by using definition – rst Jun 18 '13 at 02:45
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    Exactly, and what is your definition? What is the definition of a function? Of it having an inverse? (Note that functions can have left and/or right inverses, and having a left inverse does not mean that you have a right inverse and vice versa.) – Patrick Da Silva Jun 18 '13 at 02:51
  • Sometime we proof a result by using another result.Here if $g$ is inverse of $f$ then $f(x)=y$ iff $g(y)=x$ .source:wikipedia – rst Jun 18 '13 at 02:56
  • I agree that sometimes we do, but those other results you use are also either proved using other results but at some point one of them must be proved using a definition, and you don't seem to have one. – Patrick Da Silva Jun 18 '13 at 03:45
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Take an easy function: $f(x) = 2x + 1$.

the inverse of y = 2x + 1 is equal to $$g(x)= (x-1)/2$$

so $f(g(x))$ = $ 2(g(x)) + 1 = f(g(x)) therefore, $2((x-1)/2) + 1 = $f(g(x))$

if you multiply the first part you get $x-1 + 1 = f(g(x))$, and then by subtracting the $1 's$ you get $x = f(g(x))$.

Hope that helps.

(this isn't a proof but it shows the concept).

Also, the definition of an inverse is that if g(x) is the inverse, then f(g(x)) = x.

ROVER
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  • This is not a basic function! A basic function would be $$ f : \mathbb R \to \mathbb R \ x \mapsto 2x+1 $$ You always need three things to mention a function : its domain, its codomain and where it maps the elements of the domain. – Patrick Da Silva Jun 18 '13 at 02:38