# Hyper-extensions of Hom space

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?

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 $$|\mathrm{Hom}({^*}A,{^*}B)| = 2^{|{^*}A|} = 2^{\aleph_0^{\aleph_0}} = 2^{2^{\aleph_0}} > 2^{\aleph_0}$$.