MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\lambda$ be an infinite cardinal. Consider the Cantor cube $\Delta_\lambda = \{0,1\}^\lambda$. It is a standard fact in topology that the topological weight (= minimal cardinality for a basis) of $\Delta_\lambda$ is $\lambda$. Let $S$ be a zero-dimensional compact space of weight $\lambda$ and suppose $s\colon \Delta_\lambda\to S$ is a continuous suriection. Does there exists a closed subspace $D$ of $\Delta_\lambda$, which is homeomorphic to $S$ such that $p|_D$ is a homeomorphism?

share|cite|improve this question
The dual statement is that if $B$ is a subalgebra of cardinality $\lambda$ of the $\lambda$-generated free Boolean algebra, then the inclusion monomorphism splits, and in particular, $B$ is projective. My gut feeling is that this shouldn’t hold in general. – Emil Jeřábek Dec 4 '12 at 16:30
Are your two functions $s$ and $p$ the same function? – Ramiro de la Vega Dec 4 '12 at 18:15
up vote 5 down vote accepted

The following construction is due to Pashenkov (see "Extensions of compact spaces", Soviet Math. Dokl., 1974):

Let $X=2^\omega$ and $Z=X \times 2^X$ with product topologies everywhere. Define an equivalence relation on $Z$ by $$(x_1,y_1) \sim (x_2,y_2) \Longleftrightarrow x_1=x_2 \land y_1(x)=y_2(x) \mbox{ for all } x \in X \setminus \{x_1\}.$$

Let $\hat{Z}$ be the quotient space $Z / \sim$ and let $s: Z \to \hat{Z}$ denote the quotient map. Among other things, Pashenkov shows that $\hat{Z}$ is zero-dimensional of weight $2^{\aleph_0}$ and that $s$ is an irreducible map (i.e. there is no proper closed subset of $Z$ on which $s$ is still onto $\hat{Z}$).

Thus we get a counterexample to your question by taking $\lambda=2^{\aleph_0}$, $S=\hat{Z}$ and noting that $Z$ is homeomorphic to $2^\lambda$.

share|cite|improve this answer
This is very clever, thank you. By the way, do you think is there any name for the following property (?) of a compact space: $X$ has (?) if for every surjection $s\colon X\to X$ there is a copy $Y$ of $X$ such that $s|_Y$ is injective (a homeomorphism onto its image). – Bojan Kwitek Dec 5 '12 at 12:15

I think there is an error in the question. For this to be right, $S$ must necessarily have a subspace homeomorphic to $\Delta_\lambda$. That's not true in simple cases like $\lambda = \omega$, $S$ a convergent sequence.

share|cite|improve this answer
Of course, it was a typo. Sorry for this. I want obviously $D$ homeomorphic to $S$. – Bojan Kwitek Dec 4 '12 at 15:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.