# Contractible $T_2$-space without subspace homeomorphic to it

This is an update to an older question.

Is there a contractible $$T_2$$-space $$(X,\tau)$$ on more than $$1$$ point such that no proper subspace of $$X$$ is homeomorphic to $$X$$?

• Just a remark: no example can be provided in the category of compact topological manifolds of dimension $>0$. In fact, compact manifolds with boundary are never co-hopfian. On the other hand, compact manifolds without boundary are always co-hopfian, but they are never contractible (for instance, because the top cohomology with $\mathbb{Z}_2$-coefficients does not vanish). – Francesco Polizzi Feb 21 at 7:24
• See K VARADARAJAN Hopfian and co-hopfian objects (1992) core.ac.uk/download/pdf/13295642.pdf – Francesco Polizzi Feb 21 at 7:24
• @FrancescoPolizzi's reference: article page and MSN. – LSpice Mar 8 at 21:30

An example of such contractible (compact metrizable) space can be constructed as follows.

Let $$K$$ be the Cook continuum. It has the property that any continuous map $$f:C\to K$$ defined on a subcontinuum $$C\subset K$$ is either constant or the identity inclusion. Take any countable family $$(K_{n,m})_{n,m\in\omega}$$ of pairwise disjoint non-degenerate subcontinua of $$K$$. For every $$n,m\in\omega$$ let $$\hat K_{n,m}=K_{n,m}\times [0,1]/(K_{n,m}\times\{0\})$$ be the cone over $$K_{n,m}$$ with vertex $$v_{n,m}$$.

Let $$X_0=\hat K_{0,0}$$. For every $$n\in\mathbb N$$ we shall construct a compact metrizable contractible space $$X_n$$ that contain the space $$X_{n-1}$$ as a deformation retract. Assume that for some $$n\in\mathbb N$$ a space $$X_{n-1}$$ has been contructed. Fix any function $$f_n:\omega\to X_{n-1}$$ with dense image $$f_n(\omega)$$ in $$X_{n-1}$$. Let $$X_n$$ be the quotient space of the disjoint union $$X_{n-1}\cup\bigcup_{m\in\omega}\hat K_{n,m}$$ by the equivalence relation that identifies the vertex $$v_{n,m}$$ of each cone $$\hat K_{n,m}$$ with the point $$f_n(m)$$. The space $$X_n$$ has a natural compact metrizable topology such that $$X_{n-1}$$ is a reformation retract of $$X_n$$. This compeletes the inductive step.

Then the inverse limit $$X=\varprojlim X_n$$ is a contractible compact Hausdorff space such that every proper subspace $$P$$ of $$X$$ is not homeomorphic to $$X$$. This is because $$P$$ is either non-compact or not dense in $$X$$ and hence contains no subset homeomorphic to the continuum $$K_{n,m}$$ for suitable (large) numbers $$n,m$$.

• I don't see why you take the tree to be rooted, the root plays no role (ah I see: just to define its boundary as the set of branches, even if the resulting compactification does not depend on the choice of root). Also, you need to assume, in addition to your assumption, that $T$ equals the convex hull of its subset of branching points. Otherwise you could have a infinite branch and the resulting tree would be non-cohopfian. – YCor Mar 8 at 17:44
• Take a tree as you think of, and glue an "isolated" geodesic ray to the root (assuming that this does not destroy 2). If you don't know what a convex hull is, my additional condition is equivalent to the condition that the set of vertices of degree 2, endowed with the induced graph structure, has no infinite component. – YCor Mar 8 at 17:48
• Ah, I would never have thought of a vertex of degree as a branching point! In any case the simplest is to assume that every vertex has degree $\ge 3$. – YCor Mar 8 at 17:51
• It seems to me that many easily defined trees of the described type produce (induce) space which does have a proper homeomorphic subspace. I feel that it is still an annoying challenge to define one which does not (possibly, there is no such tree of the given type). The "secret" of counterexamples is to have a different root. – Wlod AA Mar 8 at 17:51
• @WlodAA Perfect, we agree. – YCor Mar 8 at 21:01