We know that strictly monotone functions are bijective (https://proofwiki.org/wiki/Strictly_Monotone_Function_is_Bijective). Here the set of definition is freely chosen, i.e., there is no assumption to be closed. Another result require that the interval of defiition must be closed and the image also must be closed (https://proofwiki.org/wiki/Continuous_Function_on_Closed_Interval_is_Bijective_iff_Strictly_Monotone).
Now, I have a function $$f(x)=g(x)/h(x)$$ defined on an interval $(α,2]$ where $h(α)=0$ and $lim_{x→α}f(x)=+∞$. Assuming that $f$ is continuous and strictely decreasing on $(α,2]$. Then $f$ is a bijection from $[w,2]$ to $[f(2),f(w)]$ for ll $w$ in $(α,2]$.
My question is: Can we deduce that $f$ is a bijection from the open interval $(α,2]$ to the open interval $[f(2),+∞)$.
Yes. Hint: injectivity follows from being stricly decreasing, surjectivity follows from the Darboux property.
Request for details has been made... OK.
Let $x, y \in (\alpha, 2]$ and $x \neq y$. Then either $x < y$ or $x > y$. From the fact that $f$ is strictly decreasing, in the first case we have $f(x) < f(y)$, in the second: $f(x) > f(y)$. Either way $f(x) \neq f(y)$. Hence $f$ is injective.
Let $a \in [f(2), \infty)$ be arbitrary. Since $\displaystyle \lim_{x \to \alpha^+} f(x) = \infty$, there is $\beta \in (\alpha, 2]$ such that $a \leqslant f(\beta)$. So we have $f(2) \leqslant a \leqslant f(\beta)$. Since $f$ is continuous, it has the Darboux property, i.e. there is $\gamma \in [\beta, 2]$ such that $f(\gamma) = a$. Hence $f$ is surjective.
Therefore $f$ is bijective as it is both injective and surjective.