Previous: [[Quotients of free groups]] In algebra and algebraic topology, it's often very illuminating to use the language of universal properties and categories. Historically, algebraic topology is where all this started in the first place. In these posts I'll try to be gentle about it, and introduce things slowly. Today isn't super complicated; the *good* thing about universal properties is that they are often very obvious. Maps (functions) between sets are not super complicated: we can send any element to any other element. Maps between groups (homomorphisms) are a lot more restrictive, because $ab$ must be mapped to $\phi(a) \phi(b)$ in the codomain group under the homomorphism $\phi$. However, there is almost a way to make homomorphisms less restrictive: for any group $G$, take a generating set $S$ and find out all homomorphic images of elements in $S$. From this we can construct the image of any element in $G$ because $S$ generated $G$. This is almost how the image of a linear map on a basis is enough to reconstruct the linear map^[this analogy is even more apt than you think]. The main issue is, of course, that there might be relations $R$ on the presentation $G = \langle S|R\rangle$ and so we cannot arbitrarily set $\phi(s)$ for $s \in S$; they must obey the equations from $R$. For instance, if one of the relations is $x^n = e$, then we must send $x$ to an element in $H$ whose order divides $n$ so that we satisfy the equation $\phi(x)^n = e_H$ Of course, the whole point of free groups was that there are no relations! So, if we have $F(S)$ the free group on $S$ and some map $\phi: F(S) \to H$ for some group $H$, then we can just set $\phi(s)$ to anything we want for every $s\in S$ and get a valid group homomorphism out of it! This is the spirit of the universal property of the free group. Formally, let $S$ be any set and any group. Then the universal property of the free group is that any *function* (of sets) $f: S \to H$ extends *uniquely* to a *homomorphism* $\phi_f: F(S) \to H$ such that when we evaluate $\phi_f$ on the elements $s$ of $S$ sitting inside $F(S)$, they give us exactly $f(s)$^[the map $\phi_f$ is actually referred to as $F(f)$, we'll see why later]. If you've seen commutative diagrams before, this is precisely the statement that the following diagram commutes (where $\iota$ is the inclusion of the elements of $S$ in $F(S)$) <iframe class="quiver-embed" src="https://q.uiver.app/#q=WzAsMyxbMCwwLCJTIixbMCwwLDEwMCwxXV0sWzAsMSwiRihTKSIsWzAsMCwxMDAsMV1dLFsxLDAsIkgiLFswLDAsMTAwLDFdXSxbMCwyLCJmIiwxLHsiY29sb3VyIjpbMCwwLDEwMF19LFswLDAsMTAwLDFdXSxbMCwxLCJcXGlvdGEiLDEseyJjb2xvdXIiOlswLDAsMTAwXX0sWzAsMCwxMDAsMV1dLFsxLDIsIlxccGhpX0YiLDEseyJjb2xvdXIiOlswLDAsMTAwXX0sWzAsMCwxMDAsMV1dXQ==&embed" width="310" height="304" style="border-radius: 8px; border: none;"></iframe> We can see this in action concretely. If $S = \{s_1, s_2, \ldots, s_m\}$ then all the map $f$ tells us is the values of $\phi_f$ on the elements of $S$ sitting in $F(S)$ via $\phi_f(s_i) = f(s_i)$. But since this is a generating set, we know by construction that the general element of $F(S)$ looks like $s_{i_1}^{p_1}s_{i_2}^{p_2}\ldots s_{i_n}^{p_n}$ in reduced form. Then the only possible map that extends $f$ has to be $\phi_f(s_{i_1}^{p_1}s_{i_2}^{p_2}\cdots s_{i_n}^{p_n}) = f(s_{i_1})^{p_1}f(s_{i_2})^{p_2}\cdots f(s_{i_n})^{p_n}$ and we know that there is no other equation that our homomorphism has to satisfy because there are no other relations. Said another way, because there are no relations, there is a *unique* way of writing every element of $F(S)$ in reduced form, and this is precisely what gives us this universal property. I'll talk more about category theory later, but for those who know a little bit of category theory, this universal property is an example of the **free-forgetful** adjunction. If $\mathcal S$ is the category of sets and functions, $\mathcal G$ the category of groups and homomorphisms, then it turns out that $F$ is a functor from $\mathcal G$ to $\mathcal S$^[which is why we write $\phi_f$ as $F(f)$]. We also have the functor $A$ (nonstandard notation) from $\mathcal G$ to $\mathcal S$ that "forgets" the group structure and assigns to a group its underlying set. Then the map $f$ from $S$ to $H$ is actually a morphism in $\mathcal S$ from $S$ to $A(H)$. Then our above statement says that this can be identified uniquely with a morphism in $\mathcal G$ from $F(S)$ to $H$. In category-theoretic language this is written as $\operatorname{Hom}_{\mathcal S}(S, A(H)) \cong \operatorname{Hom}_{\mathcal G}(F(S), H)$ and we say that $A$ and $H$ are **adjoint functors,** with $F$ being left-adjoint to $A$ and $A$ being right-adjoint to $F$. If you recall the analogy I gave earlier about vector spaces and bases, that was not a coincidence: In the category $\mathcal V$ of vector spaces and linear maps, the free vector space functor sending a set to all "formal linear combinations" of elements of that set is the left-adjoint to the forgetful functor from $\mathcal V$ to $\mathcal S$. Abstract nonsense aside, this universal property gives us a very slick way of showing that every group is the quotient of a free group. Recall that the way we did this was to generate $G$ by throwing in every element as a generator and every entry in the Cayley table as a relation. The first half of this can be thought of as the free group $F(G)$ ^[technically $F(A(G))$ where $A$ is the forgetful functor; this is just a type-conversion so we won't worry about this]. Now, there is an obvious map from $G$ to $G$: the identity map $\iota(g) = g$. Then by the universal property, there is a unique map $F(\iota)$ extending $\iota$ from $F(G)$ to $G$, and by the first isomorphism theorem, $G \cong F(G)/\ker (F(\iota))$ Somehow the Cayley table of $G$ is automatically captured in this induced homomorphism. If you're not super familiar with the universal property stuff yet, it's a cool exercise to try and convince yourself that this is exactly what you get by zeroing out the Cayley table of a group. Next: [[A brief intro to Cayley graphs]]