What is the origin/motivation for the adjective "free" in the term "free object"?
Does it refer to them coming "for free" (as being constructed from a set in a straight-forward manner) or does it refer some type of freedom they enjoy?
[Math] Why are free objects “free”
ct.category-theoryho.history-overviewterminology
Related Solutions
After my article was published, John Harper sent me email and said that when he was a graduate student back in the 1960s, he personally asked Leray about the term "spectral" and in particular asked whether it had something to do with the spectrum of an operator. Leray began his reply by saying, "Non"; unfortunately, before he could continue, some professors approached and interrupted the conversation.
This is perhaps some weak evidence that the spectrum of an operator is not what Leray had in mind, but unfortunately gives us no more positive information about what he did have in mind.
I predict that someone such as Steve Lack or Mike Shulman will tell you about the existence of (co)limits in Mon, and they'll do it better than I would, so instead I'll address a question in the last paragraph: do $M(0)$, $M(1)$ and $M(0 \to 1)$ tell you much about the rest of $M$? The answer is basically no.
To see this -- and to understand monads -- it's helpful to observe that if $M$ is regarded as an algebraic theory then $M(n)$ is the set of words in $n$ letters, or equivalently $n$-ary operations in the theory. For example, if $M$ is the monad for groups then $M(n)$ is the set of words-in-the-group-theory-sense in $n$ letters, which are the same as the $n$-ary operations in the theory of groups. For example, $x^3 y^2 x^{-1}$ is a typical word in two letters, and $(x, y) \mapsto x^3 y^2 x^{-1}$ is a typical binary operation (way of turning a pair of elements of a group into a single element). Similarly, if $M$ is the monad for rings then $M(n)$ is the set of polynomials over $\mathbb{Z}$ in $n$ variables, which are the $n$-ary operations in the theory of rings.
(Personally I'm happy to think of any monad on Set as an algebraic theory, although others prefer a more restrictive definition of theory. Anyway, this point of view doesn't matter in what follows.)
In particular, $M(0)$ is the set of nullary operations, or constants, in the theory, and $M(1)$ is the set of unary operations. Now let's consider on the one hand the identity monad $I$, and on the other the monad $M$ corresponding to the theory generated by a single binary operation $\cdot$ subject to the equation $x\cdot x = x$. (Or if you want a more familiar but more complicated example, take $M$ to be the monad for [not necessarily bounded] lattices or semilattices; the point is that $x \vee x = x$.) We then have $I(0) = 0 = M(0)$ and $I(1) = 1 = M(1)$, hence also $$ I(0 \to 1) = (0 \to 1) = M(0 \to 1). $$ But $I$ and $M$ are very different monads.
On the positive side, I think you might be interested in the following result; it seems in the spirit of your question. Suppose there exists an $M$-algebra with more than one element. Then:
- $M$ reflects isomorphisms
- $M$ is faithful
- the unit $\eta$ of $M$ is monic (i.e. the free $M$-algebra on a set of generators contains the generators as a subset -- it doesn't identify any of them).
What's more, all but two monads on Set have this property that there exists an algebra with more than one element. One of the exceptions is the monad $M$ with $M(A) = 1$ for all sets $A$; it's the theory generated by a single constant $e$ and the equation $x = e$. The other is the monad $M$ with $M(A) = 1$ for all nonempty sets $A$ and $M(0) = 0$; that's the theory generated by no operations and the equation $x = y$.
Edit David asked where a proof of this "positive" result could be found. I learned it from a Cambridge Part III problem sheet by Peter Johnstone, and it's probably out there somewhere in the literature (e.g. maybe in Johnstone's Sketches of an Elephant). But since consulting the literature means getting up from the couch, I'll type out a proof instead.
So, let $M = (M, \eta, \mu)$ be a monad on Set such that there exists at least one $M$-algebra with more than one element. Write $F: \mathrm{Set} \to \mathrm{Set}^M$ for the free algebra functor, and $U: \mathrm{Set}^M \to \mathrm{Set}$ for the underlying set functor; thus, $M = UF$.
First we show that $\eta: id \to M$ is monic. This means for each set $A$ and each $a, b \in A$ with $a \neq b$, we have $\eta_A(a) \neq \eta_A(b)$. By the universal property of $\eta_A$, this is equivalent to the assertion that for some $M$-algebra $X$ and map $f: A \to U(X)$ of sets, we have $f(a) \neq f(b)$. (I'd like to draw a commutative triangle to illustrate this.) Choose any $M$-algebra $X$ with more than one element: then such an $f$ can indeed be constructed.
Next we show that $F$ is faithful. It will follow that $M = UF$ is faithful, since $U$ (being monadic) is also faithful. Faithfulness of $F$ is in fact equivalent to each $\eta_A$ being monic, by general properties of adjunctions; I think that's in Categories for the Working Mathematician. Anyway, the proof is that if $f, g: A \to B$ are maps of sets with $Ff = Fg$ then $\overline{Ff} = \overline{Fg}$ (where the bar indicates transpose); but $\overline{Ff} = \eta_B \circ f: A \to UF(B)$ and similarly for $g$, and $\eta_B$ is monic, so $f = g$.
Finally we show that $F$ reflects isos. It will follows that $M = UF$ reflects isos, since $U$ (being monadic) also reflects isos. A faithful functor always reflects both epis and monos, and any map in Set that's both epi and mono is an iso. Hence any faithful functor out of Set reflects isos, and the result follows from the previous part.
Best Answer
Free objects were first defined* by MacLane in Duality for Groups. That paper gives "free" a curious political context, I quote from page 486:
You can find a critical discussion of this joke on "freedom versus fascism" at nForum and at MSE.
*As pointed out by Lee Mosher and Peter LeFanu Lumsdaine, free groups were introduced much earlier. MacLane's 1950 paper gave the first definition of a free object in the context of an arrow composition from category theory.