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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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    @HenningMakholm Thank you!2012-04-12

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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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    @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.2012-04-13