Is a bijective morphism between metric spaces necessarily an isomorphism

Question

Does the inverse morphism for a bijective isometry necessarily preserve the metric or should the preservation of the metric for the inverse morphism be stated seperately? To make myself clear my question is that does the inverse morphism in metric spaces automatically preserve the distance (anologus to the case of algebraic structures that isomorphism is a bijective homomorphism) or is the situation like e.g. toplogical spaces that the continuity should be stated seperately for the invesre function in homeomorphisms?

EDIT: In short, the OP asks "If $f:X\to Y$ is a bijective map between metric spaces such that $$d_Y(f(x_1),f(x_2))=d_X(x_1,x_2) \; \; \; \forall x_1,x_2 \in X$$ then is the inverse map $f^{-1}:Y\to X$ also distance preserving i.e. do we have $$d_X(f^{-1}(y_1),f^{-1}(y_2))=d_Y(y_1,y_2) \; \; \; \forall y_1,y_2 \in Y?$$

Have you tried a continuous bijection that is not bicontinuous? — Jacob Wakem, Jul 14 '17 at 18:44
@alephnull I'm not sure what is the relation of trying this to my question — sanaz mat, Jul 14 '17 at 18:46
Isn't a "metric space morphism" just a continuous function from the induced topologies? — Jacob Wakem, Jul 14 '17 at 18:48
@sanazmat I have added a short summary of the question that (I think!) you are asking. I hope you don't mind and also that it is as you intended! — user350031, Jul 14 '17 at 18:54

score 7 · Accepted Answer · answered Jul 14 '17 at 19:00

Suppose $f:X\to Y$ is a bijective distance preserving map between metric spaces.

Let $g:Y\to X$ be the (set theoretic) inverse of $f.$

Take any $y_1,y_2 \in Y.$

Since $f$ is surjective, there exists $x_1,x_2 \in X$ such that $y_i=f(x_i)$ for $i=1,2.$

By the distance preserving property of $f,$ we have

$$d_X(x_1,x_2)=d_Y(f(x_1),f(x_2))$$

which we may rewrite as

$$d_X(g(y_1),g(y_2))=d_Y(y_1,y_2).$$

Since $y_1,y_2$ were arbitrary, we have proved that $g$ is distance preserving.

Hope this clears things up! :)

Thank you so much. Yes, completely crystal clear. – sanaz mat Jul 14 '17 at 19:04 — sanaz mat, Jul 14 '17 at 19:04

score 0 · Answer 2 · answered Jul 14 '17 at 18:45

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A metric space isomorphism is an isomorphism on the induced topologies is a bicontinuous bijection. It is possible to have a continuous bijection (at least in general topologies, not sure about metric spaces) that is not bicontinuous.

answered Jul 14 '17 at 18:45

Jacob Wakem

2,382

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I think that "morphisms of metric spaces" are isometries (i.e. distance preserving maps) not just continuous maps. Hence an isomorphism of metric spaces is an isometry $f:X\to Y$ for which there exists another isometry $g:Y\to X$ such that $f\circ g = \mathrm{id}_Y$ and $g\circ f = \mathrm{id}_X.$ – user350031 Jul 14 '17 at 18:50
@user350031 Oh I see. Yeah I agree with you. My answer was bad. – Jacob Wakem Jul 14 '17 at 18:53
@sanazmat Yes it does, but not the other way around. Thus the distinction. – Jacob Wakem Jul 14 '17 at 18:57
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I've also seen morphisms of the category of metric spaces to be non-espanding maps do $d_Y(f(x), f(x')) \le d_X(x,x')$ for $f: X \to Y$. at least we have constant maps (and products e.g. because the projections will be such maps). – Henno Brandsma Jul 14 '17 at 20:09

Is a bijective morphism between metric spaces necessarily an isomorphism

2 Answers2