If $\mu$ is an infinite cardinal, there is a function $f:\mu\longrightarrow\mu$ such that $f^{-1}[\alpha]$ is cofinal in $\mu$ for all $\alpha\in\mu$.

Question

In the proof of Proposition 4.3. (p. 36) of the revised edition of Almost Free Modules by P.C. Eklof and A.H. Mekler, it is said that being $\mu$ an infinite cardinal, since $\mu\cdot\mu=\mu$, there is a function $f:\mu\longrightarrow\mu$ such that for all $\alpha\in\mu$, $f^{-1}[\alpha]$ is cofinal in $\mu$. I don't see why. Could someone tell me?

Answer 1

Prove the following lemma:

Let $\alpha$ be an infinite cardinal. If $A\subseteq\alpha$ and $|A|=\alpha$, then $A$ is cofinal in $\alpha$.

Now, since there is a bijection $g\colon\mu\to\mu^2$ define $f(\alpha)=\pi_1(g(\alpha))$, where $\pi_1$ is the projection on the first coordinate.

Answer 2

Fix a bijection $g:\mu\times \mu\to\mu$, then the range of $g$ over $\{\alpha\}\times \mu$ for any $\alpha\in\mu$ is cofinal in $\mu$: suppose it wasn't, then $g[\{\alpha\}\times\mu]\subset \lambda$ for some cardinal $\lambda<\mu$, but this is impossible since $\big|g[\{\alpha\}\times \mu]\big|=|\mu|>|\lambda|$

So to formulate $f:\mu\to\mu$, send any $\beta\in\mu$ to the $\alpha$ such that $g(\alpha,\gamma)=\beta$ for some $\gamma$.