# Free group and universal property

I'm trying to understand universal properties. An example is the definition of a free group (as I understand it so far):

Revised definition:

A free group $F_S$ over a set $S$ is a pair $(g,F_S)$ that satisfies the (universal) property that if $G$ is a group and $f: S \to G$ is a map then there exists a unique homomorphism $\varphi : F_S \to G$ such that $\varphi \circ g = f$.

(What I had written before: If $S$ is a set and $G$ is a group and $f: S \to G$ is an arbitrary map then the free group over $S$ is the pair $(g,F_S)$ that satisfies (the universal property) that there exists a unique homomorphism $\varphi : F_S \to G$ such that $\varphi \circ g = f$.)

Is the map $g: S \to F_S$ required to be the inclusion or can it be an arbitrary map?

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Let $s_1$, $s_2\in S$, and $f: S \to G$ with $f(s_1) \ne f(s_2)$. By the universal property you get $g(s_1) \ne g(s_2)$. So $g$ is one-to-one and wlog we can make it an inclusion. –  martini Apr 12 '12 at 12:25
Your statement of the definition is not quite right, because it makes it look like the free group can depend on $G$ and $f$. But in order for something to be a free group it has to work, simultaneously, with every $(G,f)$. –  Henning Makholm Apr 12 '12 at 12:29
Also, note that universal properties define things only up to isomorphism. So your emphasized "the" in "the pair $(g,F_S)$" is somewhat misleading because there are many pairs that satisfy the propery -- only they all happen to be isomorphic. –  Henning Makholm Apr 12 '12 at 12:32
@ClarkKent By identifying the group and its image, which are isomorphic by the First Isomorphism theorem –  M Turgeon Apr 12 '12 at 13:14
@ClarkKent You are right. But then, identify the set with its image. Since the map is injective, it is a bijection onto its image –  M Turgeon Apr 12 '12 at 13:23
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## 1 Answer

The universal property implies that the map must be a one-to-one set-theoretic map.

To see this, let $a,b\in S$ be such that $g(a)=g(b)$. Let $G$ be a nontrivial group (e.g., $G=C_2$, the cyclic group of order $2$) and let $g\in G$ be a nontrivial element. Let $f\colon S\to G$ be defined by $$f(s) = \left\{\begin{array}{ll} 1 & \text{if }s\neq a,\\ g &\text{if }s=a. \end{array}\right.$$ By the universal property, there exists a group homomorphism $\varphi\colon F\to G$ such that $f=\varphi\circ g$. In particular, $f(b) = \varphi(g(b)) = \varphi(g(a)) = f(a) = g$, hence $b=a$ (since the only element of $S$ that is mapped to $g$ by $f$ is $a$).

Therefore, $g$ is one-to-one.

Once you know it is one-to-one, you may replace $S$ with $g(S)$ and consider it to be the inclusion, since the universal property also gives:

Theorem. Let $S$ and $T$ be sets, and let $f\colon S\to T$ be a bijection. If $(g,F_S)$ and $(h,F_T)$ are free groups on $S$ and on $T$, then $f$ induces a unique isomorphism $\Phi\colon F_S\to F_T$ such that $\Phi\circ g = h$ and $\Phi(g(s)) = h(f(s))$ for all $s\in S$.

Proof. Use the universal property of $(g,F_S)$ with $h\circ f$ to obtain $\Phi$. Then use the universal property of $(h,F_T)$ with $g\circ f^{-1}$ to obtain a map $\Psi$. Finally, use the uniqueness clause of the definition to prove that $\Phi\circ\Psi$ and $\Psi\circ\Phi$ are the corresponding identity morphisms. $\Box$

So we can replace a free group on $S$ $(g,F_S)$ with the free group $(\iota,F_{g(S)})$ which is free on $g(S)$, and which is canonically isomorphic to $(g,F_S)$.

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Thank you! .... –  Matt N. Apr 13 '12 at 8:26
Can't I do it without uniqueness? You write "...use the uniqueness clause of the definition to prove that...". But if I drop the uniqueness and just get a homomorphism $\phi : F_S \to F_T$ and a homo. $\psi$ the other way such that $\phi g = hf$ and $\psi h = gf^{-1}$ then I get that $\phi \psi$ and $\psi \phi$ are the identity maps and hence mutual two sided inverses so $F_S \cong F_T$. What am I missing? –  Matt N. Apr 13 '12 at 8:58
And I have another question: why can't I speak of "the" free group over $S$ instead of a free group? In the comments to the question HenningMackholm tells me that I can't use "the". But I thought this unique up to unique isomorphism thing was exactly what lets me do this. –  Matt N. Apr 13 '12 at 9:10
@ClarkKent: How do you get that $\phi\psi$ and $\psi\phi$ are the identity map without the uniqueness clause? You get $\psi\phi g = \psi h f = gf^{-1}f = g$, but $\psi\phi g = g$ does not imply $\psi\phi=1$! $g$ is one-to-one, which means it can be cancelled on the left, but not on the right. You can have $kg = g$ with $k\neq\mathrm{id}$. –  Arturo Magidin Apr 13 '12 at 15:50
@ClarkKent: Technically, you can't speak about "the" free group over $S$ because the object is only defined up to isomorphism. The use of the definite article suggests uniqueness, period. Later, one uses "the free group" to mean "any free group, which is defined up to unique isomorphism." But technically, you shouldn't until you establish this. And certainly, in the construction, you should not use "the" because there are many ways to construct free groups. As to your second question: no, initial objects are not unique, they are only unique up to unique isomorphism. –  Arturo Magidin Apr 13 '12 at 15:53
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