When is the class of functions between sets a set? I'm reading the paper 'The big fundamental group, big Hawaiian earrings and the big free groups'. The authors state that the class of homotopy equivalences of loops in the space he dubs as the big Hawaiian earrings is not a set. I'm not really certain if what he is saying is true. 
I think my question amounts to asking is the class of functions from a set to another set not always a set?
The construction he gives of the big Hawaiian earrings (for ordinal $\omega$) is as the one point compactification of the set of disjoint open intervals with cardinality $\omega$. He then maintains that if $\omega$ has cardinality greater than $\aleph_0$ the resulting class of equivalence classes of homotopic loops is not a group since it's not a set. Is this true? I thought the Hom class of maps between all sets was a set. I'd appreciate any insight, thanks.
 A: To answer the question, putting together points made in the comments...
There is always a set of functions between two sets, unless you are using predicative foundations, where there may be no function sets, or the set theory NF(U) which is classical (even material!) but famously has no function sets.
However, this is not what the authors of the article are considering, rather they are considering (in ordinary classical foundations) a disjoint union 
$$
\coprod_{L\in BigIntervals} Top_\ast(L,X)
$$
where each $Top_\ast(L,X)$ is a set. For Hausdorff $X$, the authors state 

"We will be saved from set-theoretic difﬁculties only by the remarkable fact that all maps deﬁned on big intervals into a ﬁxed space $X$ factor through maps deﬁned on relatively small big intervals."

Thus they are considering the quotient of the above disjoint union by a relation 'factors through'. That is, for a fixed $X$, there is a set $\{L_\alpha\}_{\alpha\in I}$ of big intervals such that every $f\colon L\to X$ for an arbitrary big interval $L$ factors through some $f_\alpha\colon L_\alpha \to X$, and then $f \sim f_\alpha$. That this isn't a problem is resolved by some variant of Scott's trick, which allows for taking quotients of proper classes, or by arguing directly that the above disjoint union should be replaced by
$$
\coprod_{L_\alpha\in I} Top_\ast(L_\alpha,X).
$$
In the event that one really has a proper class, then the fact that there is a group structure (class function giving multiplication etc) on that class means that it is a group, just not a set-sized group. Some people (e.g. Conway) would call this a Group, to distinguish it; category theorists might call it a 'large group', if using the terminology of 'small' and 'large' to distinguish set-theoretic size issues.
