# find a continuous surjective map

Let $\kappa$ be a cardinal such that $cf(\kappa)=\omega$, and consider $(\kappa)^\omega$ and $(\kappa)^\kappa$ with the discrete topology. I want to define a continuous surjective map from $(\kappa)^\omega \to (\kappa)^\kappa$, but I faced a problem with surjectivity. For instance, the following map $\theta$ is continuous but not surjective: let $< x_n>$ be a cofinal set in $\kappa$, so for every $\beta \in \kappa$ there exist $x_n$ such that $\beta \leq x_n$. Consider the set $A=\{\beta: \beta \leq x_n\}$ and let $\sup A=\beta(n)$. Define $\theta:~(\alpha \kappa)^\omega \to (\alpha \kappa)^\omega$ as

$\theta(f)_\beta = f_n \mbox{ if } \beta= \beta(n)\\ 0 \mbox{ otherwise.}$

Is that correct? and how I can make it surjective?

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Out of curiosity, what is "desecrate topology"? I cannot find a definition of it. Do you perhaps mean "discrete topology"? –  Cameron Buie Jul 2 '12 at 17:25
Sorry, it is typo. You are right by desecrate topology I mean discrete topology. –  um Haitham Jul 5 '12 at 8:20
I'm not sure what you mean by $(\kappa)^\omega$. Do you mean subset of cardinality $\omega$, or sequences? –  Asaf Karagila Jul 5 '12 at 12:16
With discrete topology (in the domain), any map is continuous. Or did you mean Tychonoff? –  tomasz Jul 5 '12 at 13:16
I second Asaf's request for clarification. Also, it's consistent with ZFC to have a strict inequality of cardinals $\kappa^\omega<\kappa^\kappa$ for some $\kappa$ of cofinality $\omega$. (For example, start with a model of GCH and adjoin lots of subsets of $\aleph_1$ with countably closed forcing.) In that situation, there won's be any surjection of the sort you want (continuous or not), unless you have a really unexpected meaning for $(\kappa)^\omega$ and $(\kappa)^\kappa$. –  Andreas Blass Dec 24 '12 at 20:15