There exists a continuously differentiable bijection, $g:[a,b]\to [c,d]$ satisfying $g'(k)>0$ with $z(k)=w(g(k))$ Let $z:[a,b]\to \mathbb{C}$ and $w:[c,d]\to \mathbb{C}$ such that there exists $t(s):[c,d]\to [a,b]$ which is a continuously differentiable bijection with $t'(s)>0$ and $w(s)=z(t(s))$. 

Then I want to prove that there exists a continuously differentiable
  bijection, $g:[a,b]\to [c,d]$ satisfying $g'(k)>0$ with
  $z(k)=w(g(k))$.

I am guessing that this $g$ is $t^{-1}$ because $w(t^{-1}(k)=z(t(t^{-1})(k))=z(k)$ and as $t$ is a bijection, so $t^{-1}$ is also a bijection.
I don't know how to proceed after this. 
 A: You already proved the part where you constructed $g$ (that is $t^{-1}$). So if is understand your question right all we need to prove is that $t^{-1}$ is continuously differentiable and that $t^{-1}(k)>0$.
You know that $t(t^{-1}(k)) = k$ so using the chain rule we have that:
$$t'(t^{-1}(k)) {t^{-1}}'(k)=1 \quad \Leftrightarrow {t^{-1}}'(k) = \frac{1}{t'(t^{-1}(k))}>0$$
Using this we may conclude that ${t^{-1}}'(k)>0$ and continuously differentiable. 
Hope that helped.
EDIT:
Here's an argument for differentiability:
Note that (by the intermediate value theorem):
$$ 1 = \frac{t(t^{-1}(k+h))-t(t^{-1}(k))}{h} = t'(c_h) \left( \frac{t^{-1}(k+h) -t^{-1}(k)}{h}\right),$$
where $c_h$ belongs to the interval connecting $t^{-1}(k+h)$ and $t^{-1}(k)$. Note that we have that $c_h \to t^{-1}(k)$ as $h \to 0$. Thus we have that:
$$\lim_{h \to 0} \left( \frac{t^{-1}(k+h) -t^{-1}(k)}{h}\right) = \lim_{h \to 0}\frac{1}{t'(c_h)} = \frac{1}{t'(t^{-1}(k)}$$
Where we have used continuity of $t'$ when passing the limit inside $t'$.
