Free group

Free groups are the free objects in $𝖦 𝗋 𝗉$ . Let $𝑆$ be a set. Then $𝔽 𝑆$ has the Group presentation $⟨ 𝑆 ⟩$ , i.e. it is the minimal completion of $𝑆$ so that it becomes a group. #m/def/group Concretely, $𝔽 𝑆$ is constructed by

Inserting the identity $𝑒$
Adding an inverse $𝑎 - 1$ for each $𝑎 \in 𝑆$
Words (expressions made of group members) are only considered equal if the group laws demand so.

Likewise for any $𝑓 \in 𝖲 𝖾 𝗍 (𝑋, 𝑌)$ there exists a unique $𝔽 𝑓 \in 𝖦 𝗋 𝗉 (𝔽 𝑋, 𝔽 𝑌)$ , which is just the homomorphic extension of mapping each single-element $𝑎 \in 𝔽 𝑋$ to the corresponding $𝑓 (𝑎) \in 𝔽 𝑌$ .

Universal property

The free group has a unique extension to a functor $𝔽 : 𝖲 𝖾 𝗍 \to 𝖦 𝗋 𝗉$ so that the natural injection becomes a natural transformation $𝜄 : i d 𝖲 𝖾 𝗍 \to | 𝔽 |$ (thus creäting a Free-forgetful adjunction). This is enabled by characterising $(𝔽 𝐴, 𝜄 𝐴)$ with the following universal property:

If $𝐺$ is a group and $𝑓 \in 𝖲 𝖾 𝗍 (𝐴, 𝐺)$ is a function there exists a unique $¯ 𝑓 \in 𝖦 𝗋 𝗉 (𝔽 𝐴, 𝐺)$ so that $¯ 𝑓 𝜄 𝐴 = 𝑓$ , i.e. the following diagram commutes

Proof

Let $⊙$ denote the product in the free group. Then for any $𝑥 \in 𝔽 𝐴$ , $𝑥 = ⨀ 𝑛 𝑖 = 1 𝜄 𝐴 (𝑎 𝑖)$ with $(𝑎 𝑖) 𝑛 𝑖 = 1 \in 𝐴$ and $𝑛 \in ℕ$ . It follows that

¯ 𝑓 (𝑥) = ¯ 𝑓 (𝑛 ⨀ 𝑖 = 1 𝜄 𝐴 (𝑎 𝑖)) = 𝑛 ⨀ 𝑖 = 1 ¯ 𝑓 𝜄 𝐴 (𝑎 𝑖) = 𝑛 ⨀ 𝑖 = 1 𝑓 (𝑎 𝑖)

So $¯ 𝑓$ is already determined by $𝑓$ . Thus $𝔽 𝐴$ fulfils the universal property. If $(𝐻, 𝑗)$ also satisfies the universal property than the following diagram commutes:

giving the required unique isomorphism.

#state/tidy | #lang/en | #SemBr