Simplicial Structures in Topology

8 I Fundamental Concepts
x ∈ U, f (x) ∈ V and, since f is continuous at x, there is an open set Ux of X such
that x ∈ Ux and f (Ux) ⊂ V. Consequently, for each x ∈ X, x ∈ Ux ⊂ U and so
U ⊂ �
Ux ⊂ U,
x∈X
that is to say, U = �
x∈X Ux; this leads to the conclusion that U is open, as a union of
open sets of X. �
The continuity of a function between two topological spaces may be characterized
in two other ways, as follows.
(I.1.9) Theorem. Let X and Y be topological spaces and f : X → Y be a function.
The following conditions are equivalent:
1. The function f is continuous.
2. For every subset U ⊂ X, f(U) ⊂ f (U).
3. For every closed subset C of Y , f −1 (C) is closed in X.
Proof. 1 ⇒ 2. Assume that f is continuous; we want to prove that f (x) ∈ f (U) for
every x ∈ U. LetVbe a neighbourhood of f (x); sincefiscontinuous, f −1 (V ) is an
open set of X which contains x ∈ U and thus, f −1 (V) ∩U �= /0 (see Lemma (I.1.1)).
2 ⇒ 3. Assuming C ⊂ Y to be closed, we want to prove that the anti-image F =
f −1 (C) is closed in X. SinceF⊂F, we have to prove that, for every x ∈ F, x ∈ F.
In fact,
f (x) ∈ f (F) ⊂ f (F) ⊂ C = C
(the last inclusion is due to the fact that f (F) ⊂ C) andso,
x ∈ f −1 (C)=F.
3 ⇒ 1. Let V be any open set of Y . It follows from condition 3 that f −1 (Y �V) is
closed in X. We now note that
f −1 (V )= f −1 (Y � (Y �V)) = f −1 (Y ) � f −1 (Y �V)=X � f −1 (Y �V)
and this last set is open in X. �
(I.1.10) Corollary. Given two topological spaces X and Y, where X = C1 ∪C2 is
the union of two subspaces C1 and C2 closed in X, let f : X → Y be a function whose
restrictions to the closed sets C1 and C2 are continuous. Then, f is continuous.
Proof. Let us write f1 = f |C1 and f2 = f |C2. For each closed set V ⊂ Y,
f −1 (V)= f −1
1
−1
(V ) ∪ f2 (V ).
By the previous theorem, f −1
−1
1 (V) and f2 (V) are closed in X; therefore, f −1 (V ) is
closed in X and, consequently, f is continuous. �
If the function f : X → Y is bijective (injective and surjective), then there exists
an inverse function f −1 : Y → X such that f −1 f = 1X (the identity function
from X onto itself) and ff −1 = 1Y ; in this case, if also f −1 is continuous, we say