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