Simplicial Structures in Topology

I.2 Categories 37
(I.2.5) Lemma. Given a group G with unity 1G, a group F(S;R), and a function of
sets θ : S → G such that, for every ω ∈ R, θ(ω)=1G, there exists a unique group
homomorphism
θ : F(S;R) → G
such that θ(s)=θ(s) for every s ∈ S. 4
Proof. Given any word s ε1
1 ...sεn
n ∈ WS,define
θ([s ε 1
1 ...sεn
n ]) := θ(s1) ε1 ...θ(sn) εn . �
(I.2.6) Theorem. The category Gr is closed by pushouts.
Proof. Given any pair of homomorphisms f : G → G1 and g: G → G2, weview
the groups G1 and G2 as
and consider the set
Gi = F(Gi;RGi ) , RGi = {(xy)1 y −1 x −1 | x,y ∈ Gi} , i = 1,2
We now define the group
and the canonic homomorphisms
R f ,g = { f (x)g(x) −1 | x ∈ G}.
G := F(G1 ∪ G2;RG 1 ∪ RG 2 ∪ R f ,g)
f : G2 → G , g: G1 → G.
Since f (x)g(x) −1 is a relation in F(G1 ∪G2) for every x ∈ G, the following diagram
commutes:
f
G ��
g
��
G2
f
G1
g
��
��
G
Let us prove the universal property. Given two group homomorphisms
such that h1 f = h2g, the function
hi : Gi → H , i = 1,2
θ : G1 ∪ G2 → H , (∀x ∈ Gi) θ(x)=hi(x) , i = 1,2
4 Warning: Here s has two meanings. As an element of the domain of θ, it is the class [s], butas
an element of the domain of θ, it is just the element s.