
24 J. Almeida
Consider the cover of K by the open balls B
δ
(s)ofradiusδ centered at points s ∈ K.By
Proposition 3.9, these balls are open in the topology of S and so, since K is compact, there
are finitely many points s
1
,...,s
n
∈ K such that
K ⊆
n
i=1
B
δ
(s
i
). (4.3)
Let f ∈ V (K, U ) ∩ End S and let
W =
n
i=1
V ({s
i
},B
δ
(f(s
i
))) ∩ End S.
Then W is an open set of End S in the pointwise convergence topology which contains f.
Hence it suffices to show that W ⊆ V (K, U ). Let g ∈ W and s ∈ K.By(4.3),there
exists i ∈{1,...,n} such that d(s, s
i
) <δ. By Lemma 4.11, we have d(g(s),g(s
i
)) <δ.On
the other hand, since g ∈ V ({s
i
},B
δ
(f(s
i
))), we obtain d(g(s
i
),f(s
i
)) <δ.Sinced is an
ultrametric, it follows that d(g(s),f(s
i
)) <δ. Finally, as f (s
i
) ∈ U since f ∈ V (K, U ), we
conclude by the choice of δ that g(s) ∈ U , which shows that g ∈ V (K, U ). 2
For the case of finitely generated relatively free profinite semigroups, the following result
was already observed in [19] as a consequence of a result from [6]. We provide here a direct
proof of the more general case without the assumption of relative freeness.
4.14 Theorem Let S be a finitely generated profinite semigroup. Then End S is a profinite
monoid under the pointwise convergence topology and the evaluation mapping is continuous.
Proof By Proposition 4.13, the pointwise convergence and compact-open topologies coincide
on End S. Hence the evaluation mapping ε :(EndS) × S → S is continuous.
Since S is totally disconnected, so is the product space S
S
and its subspace End S.
Moreover, End S is compact by Lemma 4.12. Hence, by Theorem 3.1, to prove that End S
is a profinite monoid it suffices to show that it is a topological monoid, that is that the
composition µ :(EndS) ×(End S) → End S is continuous, for which we show that, for every
convergent net (f
i
,g
i
) → (f,g)in(EndS)×(End S), we have f
i
◦g
i
→ f ◦g. For the pointwise
convergence topology the latter means f
i
(g
i
(s)) → f(g(s)) for every s ∈ S. This follows from
f
i
→ f together with g
i
(s) → g(s) since the evaluation mapping is continuous. 2
Thus, if S is a finitely generated profinite semigroup then the group Aut S of its continuous
automorphisms is a profinite group since it is a closed subgroup of End S (cf. Corollary 3.2).
In particular, the group Aut G of continuous automorphisms of a finitely generated profinite
group G is profinite for the pointwise convergence topology, a result which is useful in profinite
group theory [76]. Moreover, since one can find in [76] examples of profinite groups whose
groups of continuous automorphisms are not profinite, we see that the hypothesis that S is
finitely generated cannot be removed from Theorem 4.14.
Now, for a finite set A,
Ω
A
S is a finitely generated profinite semigroup and so End Ω
A
S is a
profinite monoid. Since
Ω
A
S is a free profinite semigroup on the generating set A, continuous
endomorphisms of
Ω
A
S are completely determined by their restrictions to A.ForΩ
n
S we
may then choose to represent an element ϕ ∈ End
Ω
n
S by the n-tuple (ϕ(x
1
),...,ϕ(x
n
))
where x
1
,...,x
n
are the component projections. When n is small, we may write x,y,z,t,...
or a,b,c,d,... instead of x
1
,x
2
,x
3
,x
4
,... respectively.