ScalingStacks

Proof. The implication $(1)\Rightarrow(2)$ follows from the fact that $X_{f}\subset f^{-1}(f(E_{1}))$. For the implication $(2)\Rightarrow(3)$, take $E_{2}=X_{f}$. For $(3)\Rightarrow(4)$, take $E_{x}=(X-\{x\})\cap(f^{-1}(f(x))\cup E_{2})$. The implication $(4)\Rightarrow(5)$ is immediate.

Let us show that $(5)\Rightarrow(1)$. Note first that an injective continuous map ${\mathbf{R}}\to{\mathbf{R}}$ is open and a homeomorphism onto its image. It follows that the implication holds when $X$ and $X^{\prime}$ are homeomorphic to ${\mathbf{R}}$ and $E_{3}=\emptyset$.

Consider now the general case. There is a finite subset $E_{1}$ of $X$ containing $E_{3}$ such that $X-E_{1}$ and $X^{\prime}-f(E_{1})$ are homeomorphic to a finite disjoint union of copies of ${\mathbf{R}}$. By the discussion above, the restriction of $f$ to a connected component of $X-E_{1}$ is open and a homeomorphism onto its image, so the same holds for $f_{|X-E_{1}}$.

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