# Topological characterization of injective metric spaces

Let $\ (X\ d)\ \,(Y\ \delta)\$ be arbitrary metric spaces. A function $\ f:X\rightarrow Y\$ is called a metric map (with respect to the given metrics $\ d\ \delta$) $\ \Leftarrow:\Rightarrow\ \forall_{p\ q\in X}\ \delta(f(p)\ f(q))\ \le\ d(p\ q)$.

Injective metric spaces were introduced in a paper by Aronszajn and Panitchpakdi, under the hyper-convex spaces name, via the binary intersection property of closed balls. Equivalently, a metric space $\ (Z\ \rho)\$ is called injective $\ \Leftarrow:\Rightarrow\$ for every metric space $\ (X\ d)\$ and arbitrary $\ Y\subseteq X,\$ and for arbitrary metric map $\ f:Y\rightarrow Z\$ (with respect to $\ \delta := d|Y\times Y$ and $\ \rho$) there exists a metric map $\ g:X\rightarrow Z\$ (with respect to metrics $\ d\ \rho$) such that $\ g|Y=f$.

PROBLEM   Characterize topologically the toplogical spaces which are homeomorphic to the injective metric spaces.

Preferably, this should be done for the class of all metric spaces. The class of separable spaces or of metric compact spaces would be great too.

In the case of $1$-dimensional compact spaces $\ X\$it is pretty obvious that they are injective $\ \Leftarrow:\Rightarrow\ X$ is an AR (i.e. absolute retract as defined by Borsuk).

Sorry, if I missed some known results (I do use Google etc, but I am terrible at searching). Please, let me know.

EDIT Under a pressure from some (just one?) careful MO participants I have edited the statement of my MO-Question. I must say that in my own opinion my old statement:

Characterize topologically injective metric spaces.

is much better. I would rather say characterize topologically closed interval, circle, Euclidean plane and sphere   than   characterize topologically topological spaces homeomorphic to closed interval, circle, Euclidean plane and sphere. Or one can also say simply the same using the symbols: $\ I\ S^1\ \mathbb R^2\ S^2$. The reason to me is both mathematical, as well as the simplicity of language.

• Just in case, don't worry about spaces which are not complete--they are never injective. A non-complete space is not a retract of its completion. – Włodzimierz Holsztyński Oct 24 '14 at 6:14
• Something's wrong with typesetting in the first paragraph. Also, don't you believe in commas? – Andrej Bauer Oct 24 '14 at 11:54
• Do you mean: "Characterize those topological spaces which are homeomorphic to an injective metric space?" or "Characterize those metric spaces which are injective, but only in terms of their topological properties?" – Andrej Bauer Oct 24 '14 at 12:06
• @WłodzimierzHolsztyński Are there some known results about projective metric space with reversing the arrows in your definition? moreover, what is a relation with "functional analysis? I s it a good idea to translate this property for (commutative) separable $C^{*}$ ? – Ali Taghavi Oct 24 '14 at 16:19
• @Ali, there are no non-trivial projective objects in the metric category of all metric spaces (only the empty space and singletons are projective). There are no non-trivial projective objects in the metric category of all bounded metric spaces. And in the metric category of all spaces of diameter $\le 1,\$ a space $\ (X\ d)\$ is projective $\ \Leftarrow:\Rightarrow\ d$ is the $\ 0\!-\!1$ metric. – Włodzimierz Holsztyński Oct 24 '14 at 17:54

This might be useful:

J. R. Isbell, Six theorems about injective metric spaces. Commentarii Mathematici Helvetici 39 (1964), 65-76.

From the intro:

Aronszajn and Panitchpakdi showed  that topologically, every injective metric space is a complete retract, and asked whether the converse is true

and then Isbell goes on to state some partial results on the converse, e.g.:

In infinite 2-dimensional polyhedra, collapsibility is sufficient and free contractibility necessary

I've put the first page of the paper here. This contains a list of the results.

• Thank you @Tom. Isbell's paper is a well known classic. Of course the equivalence of the ball binary property, metric injectivity and metric absolute retract property is a very simple property (even if it can be formulated as a theorem, why not). Of course being a an absolute (topological) retract among the metrisable spaces is more general than being homeomorphic to an injective metric space (even among the separable metric spaces, or even among finite polyhedra) these two notions are not equivalent. – Włodzimierz Holsztyński Oct 24 '14 at 16:58

[Edit: I only now correctly read the question, namely you are after a topological characterization of such spaces. Please take the answer below as a useful reference on the topic of the question.]

I am not really an expert on this, but if you look this stuff up on Wikipedia and follow the link to

Espínola, R.; Khamsi, M. A. (2001). "Introduction to hyperconvex spaces". In Kirk, W. A.; Sims, B. (Eds.). "Handbook of Metric Fixed Point Theory". Dordrecht: Kluwer Academic Publishers.

you will find there Theorem 4.2 in which it is stated that a metric space is injective if, and only if, it is hyperconvex. And there is still more in the paper. Is this what you're looking for?

• Equivalence of A&P's binary intersection property and metric injectivity is about the earliest and most fundamental; I've written in my OM-QUestion from the start: Injective metric spaces were introduced ... under the hyper-convex spaces, via a binary intersection property of closed balls. Equivalently, a metric space (Z ρ) is called injective .... Thus this much should be clear from the QUESTION statement. (will continue below) – Włodzimierz Holsztyński Oct 24 '14 at 15:44
• Thank you, @Andrej, for the reference. It is not related (directly) to the question of characterization, it is not even mentioned as long as I know (actually, I've never saw it mentioned). This E&K's text belongs to the school represented for a long time by W.A.Kirk, which is concerned with non-expanding (i.e. metric) maps of metric spaces and their fixed point property--one could say, with the difficult case of this kind of the fixed point property as opposed by the Banach fpp theorem which can be considered the easy case. – Włodzimierz Holsztyński Oct 24 '14 at 15:50