Finding the inverse of a mapping that can be defined as a function on a specific domain

Question

Let $A = \{x \in \mathbb R\mid x\geq2\}$ and $B = \{x \in \mathbb R\mid x\geq1\}$ and the function $f : A\rightarrow B$ is defined by $f(x) = x^2-4x+5$.

With this domain and codomain, the function is surjective and injective as well. Because this function is bijective an inverse can be taken, which I got to be $2+\sqrt{x-1}$ and $2-\sqrt{x-1}$. $2 + \sqrt{x-1}$ would be the inverse, but my question is that can you redefine the inverse to satisfy the conditions of the domain and codomain by just stating the inverse is $2 + \sqrt{x-1}$?

Answer 1

Note $f(x) = (x - 2)^2 + 1$. The function $g : B \to A$ given by $g(x) = 2 + \sqrt{x - 1}$ is the inverse of $f$ since $$g(f(x)) = 2 + \sqrt{f(x) - 1} = 2 + \sqrt{(x - 2)^2} = 2 + (x - 2) = x$$ and $$f(g(x)) = (g(x) - 2)^2 + 1 = (\sqrt{x - 1})^2 + 1 = (x - 1) + 1 = x$$

Answer 2

This is not an exact answer, but for intuition, this may help.
Hint. Look at the highlighted part of diagram

The function is surjective and injective.

Answer 3

The given function is in fact injective.

Proof:
Assume for $x,y\in A$ with $x\ne y$, $f(x)=f(y)$.[i.e., $x\ge2,y\ge 2 $ and $x-y\ne 0$]
Then, $$\begin{align}x^2-4x+5&=y^2-4y+5\\ \implies x^2-y^2&=4(x-y)\\ \implies x+y&=4\end{align}$$ Now $x+y=4$ and $x\ge2,y\ge 2\implies x=y=2$. Hence a contradiction.

$\therefore x\ne y\implies f(x)\ne f(y)$, i.e., $f$ is injective.