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I asked the original question, and I am providing the clarifications asked in the comments
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$\mathcal G: \mathbb R_+ \to $$\mathcal G: \mathbb R_+ \to \mathbb R_+$ is a set of strictly increasing continuous functions. If for someany $\alpha<1$$\epsilon>0$,$x\in \mathbb R_+$ and $z$, and for all $\epsilon>0, x$,$\alpha\in (0,1)$ there exists $z\leq x$ and $g\in \mathcal G$ such that $g(z)\leq \alpha g(x) \leq g(z+\epsilon)$, under what topological conditions over $\mathcal G$ would there exist a $h \in \mathcal G$$h_x \in \mathcal G$ such that $\alpha h(x)=z$$\alpha h_x(x)=h_x(z)$? Are the conditions necessary/ sufficient?

$\mathcal G: \mathbb R_+ \to $ is a set of strictly increasing continuous functions. If for some $\alpha<1$, and $z$, and for all $\epsilon>0, x$, there exists $g\in \mathcal G$ such that $g(z)\leq \alpha g(x) \leq g(z+\epsilon)$, under what topological conditions over $\mathcal G$ would there exist a $h \in \mathcal G$ such that $\alpha h(x)=z$? Are the conditions necessary/ sufficient?

$\mathcal G: \mathbb R_+ \to \mathbb R_+$ is a set of strictly increasing continuous functions. If for any $\epsilon>0$,$x\in \mathbb R_+$ and $\alpha\in (0,1)$ there exists $z\leq x$ and $g\in \mathcal G$ such that $g(z)\leq \alpha g(x) \leq g(z+\epsilon)$, under what topological conditions over $\mathcal G$ would there exist a $h_x \in \mathcal G$ such that $\alpha h_x(x)=h_x(z)$? Are the conditions necessary/ sufficient?

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Finding necessary and sufficient topological conditions

$\mathcal G: \mathbb R_+ \to $ is a set of strictly increasing continuous functions. If for some $\alpha<1$, and $z$, and for all $\epsilon>0, x$, there exists $g\in \mathcal G$ such that $g(z)\leq \alpha g(x) \leq g(z+\epsilon)$, under what topological conditions over $\mathcal G$ would there exist a $h \in \mathcal G$ such that $\alpha h(x)=z$? Are the conditions necessary/ sufficient?