# Can one argue homeomorphism via equivalence of metrics? [closed]

Let $$S^n$$ be the n-sphere with respect to $$d_2$$ metric (standart metric). Let $$C^n$$ be the n-sphere with respect to $$d_1$$ metric. Clearly we have,

$$S^n$$ $$\cong$$ $$C^n$$ (homeomorphism).

Yet, can one argue this is true because $$d_1$$ and $$d_2$$ are equivalent metrics? Would this idea generalize?

## closed as unclear what you're asking by Elliot G, Yanior Weg, YuiTo Cheng, José Carlos Santos, ShaileshMay 7 at 2:33

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• What definition of equivalence of metrics are you using? – YuiTo Cheng May 6 at 13:30
• I'm confused what the question is. You want to show that $S^n$ is homeomorphic to itself with various metrics? – Elliot G May 6 at 13:44
• @YuiToCheng AFAIK all common equivalence definitions imply topological equivalence? – freakish May 6 at 14:08
• @freakish Of course. I'm just confirming whether this question is a duplicate of Equivalent metrics determine the same topology (there are also strong equivalences of metrics, you know...) – YuiTo Cheng May 6 at 14:10
• @ErotemeObelus You define "strongly equivalent". Ordinary equivalence is much weaker. – Paul Frost May 6 at 15:36

It is not a consequence of the equivalence of metrics. The reason it that the metrics $$d_i$$ on $$\mathbb R^{n+1}$$ are induced by norms $$\lVert - \rVert_i$$ on $$\mathbb R^{n+1}$$. It is well-known that all norms on $$\mathbb R^{n+1}$$ are equivalent, i.e. generate the same topology (the Euclidean topology) on $$\mathbb R^{n+1}$$. Hence each norm $$\lVert - \rVert : \mathbb R^{n+1} \to \mathbb R$$ is a continuous function with respect to this topology, and that is all we need to know.
So let us consider arbitrary norms $$\lVert - \rVert_i$$, i.e. $$\lVert - \rVert_2$$ is not necessarily the Euclidean norm $$\sqrt{\sum_{i=1}^{n+1} x_i^2}$$ and $$\lVert - \rVert_1$$ is not necessarily the norm $$\sum_{i=1}^{n+1} \lvert x_i \rvert$$. Let $$C_i = \{ x \in \mathbb R^{n+1} \mid \lVert x \rVert_i = 1 \}$$ be the unit sphere with respect to $$\lVert - \rVert_i$$.
Define $$h_1 : C_1 \to C_2, h_1(x) = x/\lVert x \rVert_2$$ and $$h_2 : C_2 \to C_1, h_2(x) = x/\lVert x \rVert_1$$. Then $$h_2(h_1(x)) = h_2(x/\lVert x \rVert_2) = \dfrac{x/\lVert x \rVert_2}{\lVert x/\lVert x \rVert_2 \rVert_1} = \dfrac{x/\lVert x \rVert_2}{(1/\lVert x \rVert_2) \lVert x \rVert_1} = x / \lVert x \rVert_1 = x$$ since $$\lVert x \rVert_1 = 1$$ for $$x \in C_1$$. Similarly $$h_1 \circ h_2 = id$$.
To see that the equivalence of metrics is not enough, consider the metric $$d_2$$ which gives you $$S^n$$. Define $$d'_2(x,y) = \min(d_2(x,y), 1)$$. This is an equivalent metric, but $$\{ x \in \mathbb R^{n+1} \mid d'_2(0,x) = 1 \} = \{ x \in \mathbb R^{n+1} \mid \lVert x \rVert_2 \ge 1 \}$$ which is not homeomorphic to $$S^n$$.
• Yes. Equivalent norms $\lVert - \rVert_i$ on $X$ induce the same topology and the functions $\lVert - \rVert_i : X \to \mathbb R$ are continuous with respect to this topology. This is needed to see that the above functions $h_i$ are continuous. – Paul Frost May 6 at 15:31