I have a question and would appreciate a clear answer.

Firstly, I will provide an introduction regarding my understanding, and then I will ask my question.

Let's begin with the definition of a universal morphism from an object $ X $ to a functor $F$:

Let $\mathcal{C}$ and $\mathcal{D}$ be two categories. A universal morphism from object $X\in ob(\mathcal{D}) $to functor $F: \mathcal{C} \longrightarrow \mathcal{D} $ is a pair $ (A, f) $, where $A$ is an object of $\mathcal{C}$ and $ f: X \longrightarrow F(A) $ is a morphism in $\mathcal{D}$. This pair satisfies the universal property, i.e., for every $ B \in \text{ob}(\mathcal{C}) $ and any morphism $ g: X \longrightarrow F(B) $, there exists a unique map $h: A \longrightarrow B$ such that the diagram commutes: $ g = F(h) \circ f $.

Now, if $(\mathcal{C}, F) $ is a concrete category and $f $ is the canonical injection, then the universal morphism $ (A, f) $ is called a free object on a set $ X $.

On the other hand, if $\mathcal{C} $ is the category of groups $\textbf{Grp} $, then the free object reduces to the free group in the category $\textbf{Grp} $.

However, the definition of a free group is:

Free group on generators $x_1,...,x_m,x_1^{-1},...,x_m^{-1}$ is a group whose elements are words in the symbols $x_1,...,x_m,x_1^{-1},...,x_m^{-1}$ subject to the group axioms. The group operation is concatenation.

Now, my question is how to reduce a free object into a free group in the category of groups.

I know that the faithful functor $F$ of the concrete category $(\textbf{Grp}, F)$ is the forgetful functor.

  • 5
    $\begingroup$ I'm glad you specified you want a "clear answer", because here I was ready to try to answer as unclearly as I could manage... $\endgroup$ Commented Mar 16 at 17:47
  • 2
    $\begingroup$ The free group on $X$ is the image of the set $X$ under the functor $\mathsf{F}\colon \mathsf{Set}\to\mathsf{Group}$ which is the left adjoint of the forgetful underlying set functor $\mathsf{U}\colon\mathsf{Group}\to\mathsf{Set}$. I have to think how to relate it to the formalism you have set up, and don't really have the time this instant. Possibly later today. And my previous point is that it is a given that you want clear/good answers. Making a point of explicitly requesting one at the top is at best passive-aggressive, and at worst condescending and insulting. $\endgroup$ Commented Mar 16 at 17:54
  • 3
    $\begingroup$ "At best, passive-aggressive" -- sheesh. I think a more charitable explanation could be that English is not OP's native language? $\endgroup$
    – user43208
    Commented Mar 16 at 17:59
  • 1
    $\begingroup$ @user43208 English is my fourth language. $\endgroup$ Commented Mar 16 at 18:00
  • 1
    $\begingroup$ Congratulations, but that doesn't matter! My point is that I think you're being a bit harsh to this newish user. I think you could have left it at the quip in your first comment. $\endgroup$
    – user43208
    Commented Mar 16 at 18:01

4 Answers 4


Two things before we start:

  1. You are misquoting the definition of free group. What you describe is the "free group on $x_1,\ldots,x_m$", not on "$x_1,\ldots,x_m,x_1^{-1},\ldots,x_m^{-1}$". The symbols $x_1^{-1},\ldots,x_m^{-1}$ are part of the construction, but they are not part of the definition. So here, the set $X$ is going to be $X=\{x_1,\ldots,x_n\}$, and not the set $\{x_1,\ldots,x_m,x_1^{-1},\ldots,x_m^{-1}\}$.

  2. I am going to change your notation, because the use of $F$ is of course natural for "functor", but in this case it conflicts with the usual notation for "free group" and "underlying set." So instead, I am going to take the "data" functor denoted as $\mathsf{U}$, for "underlying set". Your category $\mathcal{C}$ is $\mathsf{Group}$, and $\mathcal{D}$ is $\mathsf{Set}$. So the particular instance of the "free object on $X$" here would be:

A universal morphism from the set $X$ to the "underlying set functor" $\mathsf{U}\colon\mathsf{Group}\to\mathsf{Set}$ (the forgetful functor that sends a group to its undderlying set, and a group morphism to the corresponding function between underlying sets), is a pair $(A,f)$ where $A$ is a group, and $f\colon X\to \mathsf{U}(A)$, the underlying set of $A$. This pair satisfies the universal property; i.e., for every group $B$ and every set map from $X$ to the underlying set of $B$, $g\colon X\to\mathsf{U}(B)$, there exists a unique group homomorphism $h\colon A\to B$ such that $g=\mathsf{U}(h)\circ f$.

You want to understand the "usual" explicit description of "free group on $x_1,\ldots,x_m$" in terms of this definition. (See point 1 above for what the group is free "on").

So, $X$ is going to be the set $x_1,\ldots,x_m$. The group $A$ is going to be the group you describe: the set of "reduced group words" in the alphabet $x_1,\ldots,x_m$; that is, terms of the form $$w_1^{\epsilon_1}\cdots w_k^{\epsilon_k}$$ where $w_i\in \{x_1,\ldots,x_m\}$, $\epsilon_i\in\{1,-1\}$, $k\geq0$, and where "reduced" means that there is no $i$ such that $w_i=w_{i+1}$ and $\epsilon_i=-\epsilon_{i+1}$. The group operation on $A$ is "concatenate and reduce", which means that you take the words $w_1^{\epsilon_1}\cdots w_k^{\epsilon_k}$ and $v_1^{\delta_1}\cdots v_{\ell}^{\delta_{\ell}}$, and you write the word $$w_1^{\epsilon_1}\cdots w_k^{\epsilon_k}v_{1}^{\delta_1}\cdots v_{\ell}^{\delta_{\ell}},$$ and then if there are any instances of $x_k$ followed by $x_k^{-1}$, or $x_k^{-1}$ followed by $x_k$, you remove those two letters and you check the result. This is a group, and the easiest way to verify the hard part (associativity) is to use van der Waerden's trick.

So this is the group $A$. The set map $f\colon X\to \mathsf{U}(A)$ is just the inclusion map, since each element of $X$ is itself a reduced word in $A$.

To verify that this is a universal morphism (equivalently, that $(A,f)$ is the free group on $X$) we must show that given any group $B$ and any set map $g\colon X\to B$, there is a unique group morphism $h\colon A\to B$ such that $g=\mathsf{U}(h)\circ f$. This morphism $h$ is defined as follows: given a reduced group word $w_1^{\epsilon_1}\cdots w_k^{\epsilon_k}\in A$, with $w_i\in \{x_1,\ldots,x_m\}$, $\epsilon_i\in\{1,-1\}$, we define $$h(w_1^{\epsilon_1}\cdots w_k^{\epsilon_k}) = g(w_1)^{\epsilon_1}\cdots g(w_k)^{\epsilon_k}.$$ It is then a bit of work, but not too hard, to verify that this is indeed a group homomorphism and that it satisfies $g=\mathsf{U}(h)\circ f$. Moreover, it is the unique map that will do this, because $A$ is generated by $f(X)$, and the values of any morphism $\mathfrak{h}\colon A\to B$ is completely determined by its value on the generating set $f(X)$; so if $\mathfrak{h}$ satisfies that $\mathfrak{h}(x_i) = g(x_i)$, then $\mathfrak{h}=h$.

The more common way to think about the free group on $X$ within category theory is in terms of adjoint functors. Specifically, we have the concretizing functor $\mathsf{U}\colon\mathsf{Group}\to\mathsf{Set}$, the underlying set functor, which is faithful. We define a functor $\mathsf{F}\colon \mathsf{Set}\to\mathsf{Group}$ with the following property: given a set $X$ and a group $G$, there is a natural bijection $$\mathsf{Set}(X,\mathsf{U}(G)) \cong \mathsf{Group}(\mathsf{F}(X),G).$$ The functor $\mathsf{F}$ is called the left adjoint of $\mathsf{U}$, and $\mathsf{U}$ is the right adjoint of $\mathsf{F}$; $\mathsf{F}$ is the "free group functor", that associates to each set $X$ the free group on $X$. The standard way to express the universal property is the following, which elides the functor $\mathsf{U}$:

The free group on $X$, $(\mathsf{F}(X),\iota)$, is a group $F(X)$ and a set function $\iota\colon X\to \mathsf{F}(X)$ such that for every group $G$ and every set function $g\colon X\to G$, there exists a unique group homomorphism $h\colon F(X)\to G$ such that $g=h\circ\iota$.

Throughout we are abusing notation by considering a function between groups to be the same as the corresponding function between underlying sets.

So we usually think of "free object on $X$" as the image of $X$ under a functor that is the left adjoint of a concretizing functor, rather than as an object whose image is a universal arrow in the category $\mathsf{Set}$. This gives the "free commutative ring with unity on $X$" (the polynomial ring $\mathbb{Z}[X]$); the "free vector space over $F$ on $X$" (the vector space $F^{(X)}$), the "free topological space on $X$" (the set $X$ together with the discrete topology on $X$), etc.

Another way to think about the free group on $X$, entirely within the confines of group theory (without reference to functors), is to think of "the free group on $X$" as "the most general group that contains $X$ among its elements". In that sense, the set of reduced group words can be thought of as "all group elements we could construct starting from the elements of $X$, subject only to the rules that are inherent in any group: associativity, existence of inverses, and that inverses cancel each other out." You can think of it as taking products of elements, but "leaving everything that is not obvious indicated". We know that $x$ and $x^{-1}$ multiply to the identity, but in general we do not know what $x_1$ and $x_2$ will yield when multiplied... so we just "leave it indicated" by writing $x_1x_2$. When we identify the elements of $X$ with specific elements of a group $G$ via the map $f\colon X\to G$, we can now "evaluate" these products, because now we know which elements of the known group $G$ they are, and the morphism $h$ is just the "go ahead and evaluate all the expressions left indicated" morphism.

Note that like any universal object, there is in general no such thing as "the" free group on $X$. Rather, the free group on $X$ is only determined up to (unique) isomorphism, and there are many constructions that may reify it. "The set of reduced group words on $X$" is one such construction. If you want to see two other constructions, I recommend George Bergman's An Invitation to General Algebra and Universal Constructions, which is published by Springer. There is an older version available as a PDF from his website. In particular, take a look at the extensive discussion of three ways to construct "the free group" in Chapter 3.

  • $\begingroup$ I did enjoy reading in that work by Bergman. $\endgroup$
    – i can try
    Commented Mar 16 at 21:14
  • $\begingroup$ Thank you very much for this interesting explanation. $\endgroup$ Commented Mar 16 at 23:22
  • $\begingroup$ Alright, I have a question. Do you mean to say that when you select a particular group from within a category of groups, not every group can be extended to a universal morphism?Since we pick a specific group and not arbitrary. [Arturo Magidin] (math.stackexchange.com/users/742/arturo-magidin) $\endgroup$ Commented Mar 17 at 11:59
  • $\begingroup$ On other word, is the universal morphism from a set $X$ to a functor $F$ is unique? $\endgroup$ Commented Mar 17 at 12:51
  • $\begingroup$ So, if it's not unique, then the definition of the free group we provided might not align with the general definition of a free group within the category of groups? The free group we've defined could be just one of several possible instances,and we have just prove that the one we defined is one of these possibilities. Please correct to me if Iam missing something.@Arturo Magidin $\endgroup$ Commented Mar 17 at 13:14

Many definitions in mathematics (like the one you gave of "free group") are constructions. They tell you how to build a certain object.

Other definitions (like the one you gave of "free object") are properties -- for any given mathematical object, you can check whether or not it satisfies the desired property.

The central idea of category theory is to identify properties which uniquely determine important constructions up to canonical isomorphism. The definition of "free object" applies in the setting of any concrete category, but the constructions of free objects in these categories will be different.

So, you need to show two things:

(i) The construction of free groups that you provided satisfies the desired universal property;
(ii) Any two free objects on the same set are canonically isomorphic.

Part (i) is an elementary abstract algebra exercise. Part (ii) is easy if you have already learned the Yoneda lemma -- have you?

  • $\begingroup$ You don't need the Yoneda lemma for (ii). I don't think it is even helpful. $\endgroup$
    – Rob Arthan
    Commented Mar 16 at 20:17
  • $\begingroup$ I disagree; the fact that any two objects satisfying the same universal property are isomorphic is essentially the same as the statement of the Yoneda lemma. $\endgroup$ Commented Mar 16 at 22:04
  • $\begingroup$ But the fact that any two objects satisfying the same universal property are isomorphic is (at least in this case) immediate from two applications of the universal property without any thoughts about natural transformations or conditions on the size of hom-sets. $\endgroup$
    – Rob Arthan
    Commented Mar 16 at 22:34

Maybe I dont completely understand your question but I dont think "universal morphism" is a thing.

This is the definition of a free object that Im familiar with: let $C$ be a concrete category, namely exists a faithful $U:C\to \textbf{Set}$ (Underlying set).

Given a set $X$, a free object with the basis $X$ is a pair $(A,i)$ where $A\in ob(C)$ and $i:X\to U(A)$ is an injective function, such that: for any $B\in ob(C)$ and a function $g:X\to U(B)$, exists a unique morphism $f:A\to B$ such that $g=U(f)\circ i$.

It follows from this definition that free objects whose bases are equinumerous are isomorphic. Specifically, a free object on a given set is - if exists - unique up to isomorphism, so simplistically we can call it the free object on a set $X$. In the category of groups we can explicitly build a free group on any set, so this is the free group on that set.

  • 1
    $\begingroup$ "Universal morphism" is a thing; for example, the canonical embedding $u_X: X \to UFX$ of a set $X$ (into the underlying set of the free group it generates) is universal among all functions $X \to UG$ into groups. This language is used for example in Mac Lane's Categories for the Working Mathematician (he uses e.g. the terminology "universal element" $u_X \in \hom(X, UFX)$ for the functor $\hom(X, U-)$ that is represented by $FX$). $\endgroup$
    – user43208
    Commented Mar 16 at 18:37
  • $\begingroup$ i havent ever seen this so i mean... i guess you could understand free objects without it (if the OP uses this book then maybe i wasnt much of a help tho) $\endgroup$
    – rutruttt
    Commented Mar 16 at 18:46
  • $\begingroup$ Not really meaning to pile on, but I'll add that for the purposes of discussing universal properties in a suitably general manner, the faithfulness of $U$ is unimportant, and injectivity of $i$ is similarly a red herring. The discussion centers on whether every functor of type $\hom(X, U-)$ is representable. $\endgroup$
    – user43208
    Commented Mar 16 at 23:43

Perhaps what the OP meant to say is that one way to construct a free group on a set $X = \{x_1, \ldots, x_n\}$ is to

  • Start by taking a free monoid $M$ on a set $\{x_1, \ldots, x_n, x_1^{-1}, \ldots, x_n^{-1}\}$ (where the $x_i^{-1}$ are new symbols that we adjoin to $X$), whose elements are formal "words" in letters drawn from the new set, multiplied by concatenation, and then

  • Pass to the monoid whose elements are equivalence classes of words, taking the equivalence relation to be the the least equivalence relation containing pairs $(x_i x_i^{-1}, e)$ and $(x_i^{-1} x_i, e)$ where $e$ is the empty word, and such that whenever $(a, b)$ and $(c, d)$ belong to the equivalence relation, then so does $(ac, bd)$. Equivalence classes $[a]$ are multiplied by the rule $[a] [b] = [ab]$.

The resulting monoid is a group (define $[w]^{-1}$ to be the equivalence class of the word obtained by writing the word $w$ in reverse and replacing each of its letters by its formal inverse). It takes a few lines to verify that all this works, but it's straightforward. Let's denote this group by $F(X)$. Letting $UG$ denote the underlying set of a group $G$, there is a function $X \to UF(X)$ that sends $x_i$ to $[x_i]$.

Now let $G$ be a group, and suppose we are given a function $f: X \to UG$ to the underlying set of $G$. This extends to a function

$$\{x_1, \ldots, x_n, x_1^{-1}, \ldots, x_n^{-1}\} \to G$$ by sending $x_i^{-1}$ to $f(x_i)^{-1} \in G$. This function further extends uniquely to a monoid homomorphism $\hat{f}: M \to G$, in a pretty obvious way. Finally, by passing to equivalence classes as described above, this uniquely induces a homomorphism $F(X) \to G$ between groups, defined by the rule $[w] \mapsto \hat{f}(w)$ (check this is well-defined!). Since everything here was forced, the uniqueness clause for the universal property is assured.

I've left a certain number of details to you to check, but this should give the idea.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .