Is the Subcategory of Infinite Sets a Groupoid?

Question

The following exercise is given in the text "Algebra Chapter 0" by Aluffi:

Construct a category of infinite sets and explain how it may be viewed as a full subcategory of Set.

Let $\infty\text{-Set}$ denote the category whose objects are all sets $X$ with $\vert X \vert=\infty$ with functions between them. Clearly it should be the case that $X,Y\in{Ob(\infty\text{-Set})}\Rightarrow \vert X \vert=\vert Y \vert=\infty$ for any pair of objects which would imply that $X\cong Y$ making $\infty\text{-Set}$ a groupoid. My reservations in claiming that the above category is a groupoid stems from the following: $\Bbb{R}$ and $\Bbb{Z}$ both have infinite cardinality but $\lnot(\Bbb{R} \cong \Bbb{Z}$) since their respective cardinalities are different sizes of infinity.

Finally my question: Given the information in the exercise, can I conclude that $\infty\text{-Set}$ is a Groupoid or would one need to add that the objects are restricted to countably infinite sets or something like that? Thanks.

Answer 1

You have several points of confusion.

First, $|X|=\infty$ is really just an abbreviation for "$X$ is infinite", and should not be read literally. There is no cardinal number "$\infty$", since infinite sets have many different cardinalities, and so you cannot say that if $|X|=\infty$ and $|Y|=\infty$ then $|X|=|Y|$. Really, the notation $|X|=\infty$ is usually not used except in contexts where you care mainly about finite sets, and so you want to distinguish that a set is infinite in contrast to them.

Second, to say that a category is a groupoid does not mean that any two objects are isomorphic. Rather, it means that every morphism is an isomorphism. It is possible to have a category in which any two objects are isomorphic, but not every morphism is an isomorphism (between any two objects there exists an isomorphism, but there also exist other morphisms which are not isomorphisms). So for instance, even if you restricted to the category of countably infinite sets, you would not have a groupoid, because not every function between two countably infinite sets is a bijection. It is also possible to have a groupoid in which not all objects are isomorphic, since you can have two objects with no morphisms between them at all.

Answer 2

Some thoughts:

• "Full" subcategory means that we retain all the maps we did in $\mathsf{Set}$. So for instance, the map $f : \mathbf{Z} \to \mathbf{Z}$ defined by $f(x) = 0$ is in the category. I don't need to tell you that this isn't an isomorphism.

• You wrote $|X| = |Y| = \infty$. Usually, for infinite sets, we use cardinal numbers, not the symbol $\infty$ and $|X| = |Y|$ means there is a bijection $X \leftrightarrow Y$. So we might say $|\mathbf{Z}| = \aleph_0$ and $|\mathbf{R}| = 2^{\aleph_0}$.

• If we wanted a groupoid, we could fix some cardinal number $\alpha$ (e.g. $\alpha = \aleph_0$ if we wanted countable sets) and look at all objects $X$ in $\mathsf{Set}$ with $|X| = \alpha$. We wouldn't take all of the morphisms we had in $\mathsf{Set}$ because we can't. Instead we must restrict to those morphisms which are isomorphisms. So what we would end up with isn't a full subcategory.

• We can also consider several cardinal numbers. For instance, look at the subcategory of all finite groups whose morphisms are bijections. This is a groupoid even though some objects are not isomorphic. That's okay, we just don't have maps between such objects.