How can I prove that a function is invertible, where do I get the function and the necessary data?

Question

"If g is another invertible function, then the compound function $f ∘ g$ is also invertible, and it is fulfilled.."

$$( f ∘g) ^{-1} = g^{-1} ∘ f^{-1}$$

I don't even understand where to start. With what procedures can I demonstrate it?

Answer 1

Suppose $$y=(f \circ g)(x);\;y=f(g(x))$$ apply $f^{-1}$ to both sides $$f^{-1}(y)=f^{-1}(f(g(x)))\;f^{-1}(y)=g(x)$$ Now apply $g^{-1}$ to both sides $$g^{-1}(f^{-1}(y))=g^{-1}(g(x))\;g^{-1}(f^{-1}(y))=x$$ Now swap $x$ and $y$ $$y=g^{-1}(f^{-1}(x))\;y=(g^{-1}\circ f^{-1})(x)$$

We have proved that $$(f\circ g)^{-1}=g^{-1}\circ f^{-1}$$

Answer 2

As far as your first question concerns, $f\colon X\to Y$ is invertible if (definition) $\forall y\in Y,\exists!x\in X\mid y=f(x)$. Likewise, $g\colon Y\to Z$ is invertible if $\forall z\in Z,\exists!y\in Y\mid z=g(y)$. Therefore, if both $f$ and $g$ are invertible, then $\forall z\in Z, \exists!x\in X\mid z=g(f(x))=:(gf)(x)$, which precisely means that $gf\colon X\to Z$ is invertible.

Note that from the very synthax of $f$'s invertibility ("...$\forall y\in Y,\exists!x\in X\mid y=f(x)$...") comes the existence of a map $h_f\colon Y \to X$ such that $x=h_f(y)$, and hence:

$$y=f(h_f(y)), \space\forall y\in Y \tag 1$$

The same holds for $g$ (invertible!), for some $h_g\colon Z\to Y$ such that:

$$z=g(h_g(z)), \space\forall z\in Z \tag 2$$

and for $gf$ (invertible!) for some $h_{gf}\colon Z\to X$ such that:

$$z=(gf)(h_{gf}(z)), \space\forall z\in Z \tag 3$$

But then, from $z=z$ and $(2)$ & $(3)$, we get: $g(f(h_{gf}(z)))=g(h_g(z)), \forall z\in Z$, which implies (by the injectivity of $g$): $f(h_{gf}(z))=h_g(z), \forall z\in Z$. By $(1)$, taking $y=h_g(z)$, this latter reads: $f(h_{gf}(z))=f(h_f(h_g(z))), \forall z\in Z$, and hence, by the injectivity of $f$: $h_{gf}(z)=h_f(h_g(z)), \forall z\in Z$, and finally:

$$h_{gf}=h_fh_g \tag 4$$

which is just another notation for $(gf)^{-1}=f^{-1}g^{-1}$.

Answer 3

A function $f:A \to B$ is invertible if there is a well defined function $f^{-1}: B\to A$ so that for any $x\in A$ we will have $f^{-1}(f(x)) = x$.

So if we are given $f:A\to B$ and $g:C\to A$ where both $f,g$ are invertible.

$f\circ g:C \to B$.

Now $f$ is invertible so for any $x\in C$ and $g(x) \in A$ we have $f^{-1}(f(g(x))= g(x)$.

ANd $g$ is invertible so for any $x\in C$ we have $g^{-1}(g(x)) = x$.

So $g^{-1}(f^{-1}(f(g(x))))= g^{-1}(g(x)) = x$.

And that's it. $g^{-1}\circ f^{-1}: C\to B$ so that for any $x \in C$ we have $(g^{-1}\circ f^{-1})(f\circ g(x))= g^{-1}(f^{-1}(f(g(x)))) = x$. So $f\circ g$ is invertible with $(f\circ g)^{-1} = g^{-1}\circ f^{-1}$.