We fix an ultrafilter $\mathcal{F}$ of $\mathbb{N}$ which contains the cofinite filter. Let $A,B$ be sets and ${}^{*}A,{}^{*}B$ their hyper-extensions. Then is $$ {\rm Hom}({}^{*}A,{}^{*}B) $$ equal to the hyper extension of ${\rm Hom}(A,B)$? Let $f=[f_{m}]_{m\in \mathbb{N}}$ be an element of ${}^{*}{\rm Hom}(A,B)$. Then for any $a=[a_{m}]_{m\in \mathbb{N}}$, we define $f(a)$ by $[f_{m}(a_{m})]_{m\in \mathbb{N}}\in {}^{*}B$. Perhaps this is well-defined, so we can consider a natural map $$ {}^{*}{\rm Hom}(A,B)\longrightarrow {\rm Hom}({}^{*}A,{}^{*}B). $$ My question is whether or not this natural map is bijective. If not, is this injective or surjective or neither?


1 Answer 1


The map your describe is well-defined, and it is always injective.

I think the easiest way to see this is by Łós's theorem. Consider the language with three sorts, $A$, $B$, and $H$, and a binary function symbol $\text{eval}\colon H\times A\to B$. Let $M$ be the structure where $A^M$ is the set $A$, $B^M$ is the set $B$, $H^M$ is the set $\text{Hom}(A,B)$, and $\text{eval}^M$ is the evaluation function $\text{eval}^M(f,a) = f(a)$.

Now let $M^\mathcal{F}$ be the ultrapower by $\mathcal{F}$, so $A^{M^\mathcal{F}} = {^*}A$, $B^{M^\mathcal{F}} = {^*}B$, and $H^{M^\mathcal{F}} = {^*}\mathrm{Hom}(A,B)$. The function $\text{eval}^{M^\mathcal{F}}$ is defined pointwise, so $\text{eval}^{M^\mathcal{F}}([f_m],[a_m]) = [\text{eval}^M(f_m,a_m)] = [f_m(a_m)]$. This is the function you considered in your question, and it induces (by currying) a function $h\colon {^*}\mathrm{Hom}(A,B)\to \mathrm{Hom}({^*}A,{^*}B)$.

$M$ satisfies the sentence $(\forall f\in H)(\forall g\in H)(((\forall a\in A)\,\text{eval}(f,a) = \text{eval}(g,a))\rightarrow f = g)$, so by Łós's theorem $M^{\mathcal{F}}$ satisfies this sentence too. This implies that $h$ is injective.

What about surjectivity? It is clear that $h$ is surjective when $|B| \leq 1$, since then $|{^*}B| \leq 1$, and both the domain and codomain of $h$ have size $\leq 1$. $h$ is also surjective when $A$ is finite, since then ${^*}A = A$, $B^A\cong B^n$ when $|A| = n$, and ultrapowers commute with finite cartesian products.

We would also get surjectivity if $\mathcal{F}$ was $\kappa$-complete and $|A|<\kappa$ (for the same reason: $\kappa$-complete ultrapowers commute with cartesian products of size $<\kappa$), but no non-principal ultrafilter on $\mathbb{N}$ is $\kappa$-complete for $\kappa>\aleph_0$.

Working with non-principal ultrafilters on $\mathbb{N}$, I believe $h$ always fails to be surjective when $A$ is infinite and $|B|\geq 2$.

We can see this in some cases for a simple cardinality reason: if $\mathcal{F}$ is a non-principal ultrafilter on $\mathbb{N}$, then for any infinite set $X$, $|{^*}X| = |X|^{\aleph_0}$ (note that any non-principal ultrafilter on $\mathbb{N}$ is $\aleph_0$-regular, and see Proposition 4.3.7 in Chang & Keisler). So taking $|A| = \aleph_0$ and $|B| = 2$, we have $|{^*}\mathrm{Hom}(A,B)| = (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0}$, while $|\mathrm{Hom}({^*}A,{^*}B)| = 2^{|{^*}A|} = 2^{\aleph_0^{\aleph_0}} = 2^{2^{\aleph_0}} > 2^{\aleph_0}$.


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