In this section, we shall get more practice with universal properties for various algebraic constructions. First, take the following categories:

*Set*= category of sets, with morphisms = set functions;*Grp*= category of groups, with morphisms = group homomorphisms;*Ab*= category of abelian groups, with morphisms = group homomorphisms.

Given any set *X*, we get the free group *F*(*X*) generated by *X*; we saw earlier that the injection *i* : *X* → *F*(*X*) gives rise to the following correspondence:

Mor* _{Set}*(

*X*,

*H*) = Mor

*(*

_{Grp}*F*(

*X*),

*H*) for any group

*H*,

where the LHS function *f* : *X* → *H* is obtained from *g* : *F*(*X*) → *H* by *f* = *gi*. This can be seen heuristically as follows: if we pick the word *aba*^{-2}*c*^{3} in *F*(*X*) (where *a*, *b*, *c* are elements of *X*), then *g* must naturally map it to *f*(*a*)*f*(*b*)*f*(*a*)^{-2}*f*(*c*)^{3} in *H*.

What if we wish to find an *abelian group F’*(*X*) such that:

Mor* _{Set}*(

*X*,

*H*) = Mor

*(*

_{Ab}*F’*(

*X*),

*H*) for any abelian group

*H*?

Let’s see what goes wrong if we put *F’*(*X*) = free group *F*(*X*). If *f* : *X* → *H* is any function, we can forget for now that *H* is abelian and reason as before that it must induce a group homomorphism *g* : *F*(*X*) → *H*. The problem is that *F’*(*X*) is not abelian, so the RHS is not well-defined since we’re working in the category of abelian groups there. To rectify, we let *F’*(*X*) be the abelianisation of *F*(*X*):

where we have |*X*| copies of **Z**, but we only take those elements with *finitely many non-zero terms*. This subgroup is called the **direct sum** of |*X*| copies of **Z**. E.g. the element 2*a* + 3*b* – 4*c* in *F’*(*X*) (where *a*, *b*, *c* in *X*) gets mapped to the element * f(a)^{2}f(b)^{3}f(c)^{-4} in H*.

Exercise : why do we only take the elements with finitely many non-zero terms? Thus, suppose we were greedy and take the set-theoretic product (i.e. the

direct product):what will go wrong? Hint: every f : X → H will still give F'(X) → H, but it is possible to find to different F'(X) → H which restricts to the same f.

Finally, we consider the third possibility: given a group *G*, we wish to find an abelian group *F”*(*G*) and group homomorphism *π* : *G* → *F”*(*G*) such that:

Mor* _{Grp}*(

*G*,

*H*) = Mor

*(*

_{Ab}*F”*(

*G*),

*H*) for any abelian group

*H,*

where *g* : *F”*(*G*) → *H* corresponds to *g π* :

*G*→

*H*. We’ll give you the answer straight and let you do the verification by yourself: the abelianisation

*F”*(

*G*) =

*G*

^{ab}.

Exercise : given a set X, we need a group E(X) and set-theoretic function σ : E(X) → X, such that

Mor

_{Set}(G, X) = Mor_{Grp}(G, E(X)),where g : G → E(X) corresponds to σg : G → X. Does such an E(X) exist?

**Fibred Products.** Next, we consider another case of universal properties. Let *ρ*_{1} : *G* → *K* and *ρ*_{2} : *H* → *K* be two fixed group homomorphisms. We wish to find the group *P*, together with homomorphisms *π*_{1} : *P* → *G* and *π*_{2} : *P* → *H* such that *ρ*_{1}*π*_{1} = *ρ*_{2}*π*_{2}, and

- for any group
*Q*and*σ*_{1}:*Q*→*G*and*σ*_{2}:*Q*→*H*such that*ρ*_{1}*σ*_{1}=*ρ*_{2}*σ*_{2}, there is a unique*f*:*Q*→*P*for which*π*_{1}*f*=*σ*_{1}and*π*_{2}*f*=*σ*_{2}.

We denote this group by *G* ×* _{K} H*.

Exercise : prove that the group

together with the projection maps to G and H, satisfies the universal property of the fibred product.