I am reading the notes about quasisymmetric maps by Juha Heinonen - Lectures on Analysis on Metric Space, chapter 10 and got stuck on something, which I guess is quite elementary..
We say that an embedding $f:X \to Y$ , where $X$ and $Y$ are metric spaces is quasisymmetric (QS) if there is a homeomorphism $\eta: [0, \infty) \to [0, \infty)$ so that $$ |x-a| \leq t|x-b| \text{ implies } |f(x)-f(a)| \leq \eta (t) |f(x)-f(b)| $$
for $a,b,x \in X$ and $t>0$.
It is stated that the inverse of QS embedding from its image is QS with homeomorphism $\eta'(t)= \frac{1}{\eta^{-1}(t^{-1})} $
Somehow I am failing to show it - I am sure I am missing some properties of homeomorphisms $f$ and $\eta$ and how the inequalities behave under them.
So we have to show that if $$ |x'-a'| \leq t|x'-b'|$$ this implies $$ |f^{-1}(x')-f^{-1}(a')| \leq \eta' (t) |f^{-1}(x')-f^{-1}(b')| $$ for $a',b',x' \in f(X)$ and $t>0$ and a homeomorphism $\eta'$. We have that $a'=f(a), b'=f(b), x'=f(x)$ for some $a,b,x \in X$ and so $$ |f^{-1}(x')-f^{-1}(a')|= |x-a| $$ and I am not sure how to proceed. Definitely have to use the quasisymmetry of $f$ to say something but I can't see how.
Any suggestions would be appreciated!
Assume for a contradiction that $|x'-a'|\leq t|x'-b'|$ but $|x-a|>\eta'(t)|x-b|$. Then $|x-b|<\eta^{-1}(\frac{1}{t})|x-a|$, so applying $f$ leads to $|x'-b'|<\frac{1}{t}|x'-a'|$, contradicting our assumption.