2

The definition given by Hungerford in his text Algebra is the following, in a concrete category $\mathcal{C}$, given an object $A$, a set $X$ and a map $\iota:X\to A$, we say that $A$ is free on $X$ if for any object $B$ and map $f:X\to B$ there exists a unique morphism $g:A\to B$ such that $g\circ\iota=f$.

If anything this should be called that $A$ is free on the pair $(X,\iota)$, or is it immediately clear that if $A$ is free on $(X,\iota)$, then $A$ is also free on $(X,\kappa)$? If $X$ is a subset and $\iota$ is the inclusion map, then there is no such ambiguity but he never mentions this.

spinosarus123
  • 2,612
  • 1
  • 8
  • 22
  • Obviously, giving just the set $X$ but not the map does not define the "set on which $A$ is free". Take the category of Abelian groups, and a 1-element set $X={a}$. The group $A=\mathbb Z$ is free with respect to the map $a\mapsto 1$ but not with respect to the map $a\mapsto 2$. The latter because the map $f: X\to \mathbb Z/2\mathbb Z, a\mapsto 0$ extends to a map $g:\mathbb Z\to \mathbb Z/2\mathbb Z$ in two different ways - as $n\mapsto 0$ and $n\mapsto n\pmod 2$. –  Sep 21 '21 at 11:38

1 Answers1

2

Related: definition of a free object in a category .

If the category has at least one object $B$ which, as a set, has at least two elements $a, b\in B, a\ne b$, then $\iota$ has to be an injection. This is because, for every $x,y\in X, x\ne y$, construct any map $f:X\to B$ mapping $x$ to $a$ and $y$ to $b$ and so $f(x)\ne f(y)$, which implies $\iota(x)\ne \iota(y)$.

So, for $\iota$ to not be an injection, either:

  • All objects in the category have at most one element (for an example see the above link, however, this is quite an odd case), or:
  • $\iota$ is an injection, and then you can identify $X$ with a subset of $A$.
  • What if the objects of $\mathcal{C}$ are not sets (or structures on sets)? – md2perpe Sep 21 '21 at 12:17
  • @md2perpe $\mathcal C$ is a concrete category, and so I was being slightly informal. –  Sep 21 '21 at 12:54
  • Ah, I missed that it was concrete. – md2perpe Sep 21 '21 at 14:05
  • I thought I got it that’s why I marked your answer as sufficient, but isn’t it still ambiguous? $A$ can be both free on $X$ and not free on $X$ depending on which injection I choose. – spinosarus123 Sep 21 '21 at 14:35
  • @spinosarus123 I agree with you, but I think this is largely irrelevant. A bit of sloppiness in the language doesn't do too much harm here. If you say that $\mathbb Z$ is a free Abelian group over the set ${1}$, you assume the natural embedding $1\to 1$, and you also assume some other things (e.g. that the group operation is addition of integers). We can use a more precise language, of course, but this would distract from the real problems, which are e.g. how to decide if a given object is free, over which set (of "generators"), do every two sets of generators have the same cardinality etc. –  Sep 21 '21 at 15:30