How does a unique solution for $x$ in $x=f^{-1}\left(y\right)$ imply a unique solution for $f\left(x\right)=y$

Question

I am doing the Khan Academy linear algebra course. I am around halfway through in which Sal talks about linear maps.

I am watching the video 'proof: Invertibility implies a unique solution to f(x)=y', in which he proves that for:

$x=f^{-1}\left(y\right)$

...there is only 1 unique $x$ that satisfies the equation.

He then says that because of this we can conclude that for:

$f\left(x\right)=y$

...there is only 1 unique solution.

How does the first point imply the second? This is what has been stumping me.

Answer 1

We have that $f^-1(y)=x$ and we have to show that $f(x)=y$.

Simply substitute $f^-1(y)=x$ in $f(x)=y$ to get $f(f^-1(y))=y$. Then we get that y=y, which of course is unique since we had fixed the value of y.