Definition 3.24 A map f: X → Y between two topological spaces is continuous
if the pre-image
f−1(A) = {x ∈ X | f (x) ∈ A}
of any open set A⊂ Y is open in X.
Before we continue let’s remark that the operator f−1: P(Y ) → P(X) commutes with the operations of set-complementation and union:
3.3 Continuous Maps 47
f−1(Y − A) = X − f−1(A), f−1(∪iAi) = ∪if−1(Ai).
As a result:
1. a map f: X → Y is continuous if and only if the pre-image f−1(C) of any closed set C⊂ Y is closed in X (complementation);
2. letB be a topological basis of Y . A map f : X → Y is continuous if and only if for any B ∈ B the set f−1(B) is open in X (each open set in Y is the union of elements ofB).
Lemma 3.25 Let f: X → Y be a map between two topological spaces. Then f is
continuous if and only if f( A ) ⊂ f (A) for any subset A ⊂ X.
Proof Let’s suppose f continuous; then for any A ⊂ X the subset f−1( f (A) ) is
closed and contains A. Therefore A⊂ f−1( f (A) ) and so f ( A ) ⊂ f (A). Now assume f( A ) ⊂ f (A) for any A ⊂ X. In particular for every closed set
C ⊂ Y (setting A = f−1(C))
f( f−1(C) ) ⊂ f ( f−1(C)) = C = C,
whence f−1(C) ⊂ f−1(C). But this means f−1(C) is closed. Lemma3.25implies that Definitions1.4and3.24are equivalent.
Theorem 3.26 The composite of continuous maps is continuous.
Proof Take continuous maps f: X → Y and g : Y → Z and let A ⊂ Z be open. By
the continuity of g we have that g−1(A) is open, and then f−1g−1(A)becomes open because f is continuous. To finish it suffices to note that f−1g−1(A) =
(g f )−1(A).
Definition 3.27 A map f: X → Y between topological spaces is continuous at a point x ∈ X if for any neighbourhood U of f (x) there is a neighbourhood V of x
such that f(V ) ⊂ U.
Theorem 3.28 A map f: X → Y is continuous if and only if it is continuous at
every point of X .
Proof Suppose f continuous and let U be a neighbourhood of f(x). By definition
of neighbourhood there is an open set A⊂ Y such that f (x) ∈ A ⊂ U: the open set
V = f−1(A) is a neighbourhood of x and f (V ) ⊂ U.
Conversely, suppose f is continuous at every point and let A be an open set in Y . We claim f−1(A) is a neighbourhood of all of its points. If x ∈ f−1(A) then A is a neighbourhood of f(x) and there is a neighbourhood V of x such that f (V ) ⊂ A. This amounts to saying V ⊂ f−1(A), and therefore f−1(A) is a neighbourhood
Definition 3.29 A homeomorphism is a continuous bijective map with continuous
inverse. More precisely, a continuous mapping f: X → Y is called a homeomor- phism if there exists a continuous map g: Y → X such that the composites g f and
f g are the identity maps on X and Y respectively.
Two topological spaces are said homeomorphic if there’s a homeomorphism between them.
Definition 3.30 A map f : X → Y is called:
1. open if f(A) is open in Y for any open set A of X; 2. closed if f(C) is closed in Y for any closed set C of X.
Lemma 3.31 Let f: X → Y be a continuous map. The following conditions are
equivalent:
1. f is a homeomorphism; 2. f is closed and bijective; 3. f is open and bijective.
Proof Let’s prove(1) ⇒ (2): every homeomorphism is bijective by very definition.
If g: Y → X is the inverse of f , g is continuous and for any closed subset C ⊂ X,
f(C) = g−1(C) is closed in Y .
As for(2) ⇒ (3), let A ⊂ X be open and set C = X − A. As f is bijective we have f(A) = f (X − C) = Y − f (C), so f (A) is the complement of the closed set
f(C).
At last, to show(3) ⇒ (1) suppose g : Y → X is the inverse of f . Then for any open subset A⊂ X, g−1(A) = f (A) is open and therefore g is continuous. The term real(-valued) continuous map typically refers to a continuous map
f: X → R for the Euclidean topology on R.
Exercises
3.17 Prove that constant maps are continuous, irrespective of the topologies consid-
ered.
3.18 Let T1 andT2 be given topologies on a set X . Prove that the identity map
(X, T1) → (X, T2), x → x, is continuous if and only if T1is finer thanT2. 3.19 Given f: R → R and k ∈ R write
M(k) = {x ∈ R | f (x) > k} m(k) = {x ∈ R | f (x) < k}.
Prove that f is continuous if and only if M(k) and m(k) are open for any k.
3.20 Let fi: Yi → X be a family of continuous maps with values in a given topo-
logical space X . Show that the induced map f: iYi → X is continuous, where
3.3 Continuous Maps 49 3.21 Two subsets A, B of a topological space are called adherent to each other if
(A ∩ B) ∪ (A ∩ B) = ∅.
Prove that continuous maps preserve the relation of adherence of subsets. Con- versely if f: X → Y is a map between T1 spaces (see Exercise3.12) that preserves adherence, f is continuous.
3.22 Prove that the composite of two homeomorphisms is a homeomorphism and
the inverse of a homeomorphism is a homeomorphism, so that the set Homeo(X) of homeomorphisms from a topological space X to itself is a group. Introduce, on the category2of topological spaces, the relation ‘X ∼ Y ⇐⇒ X is homeomorphic to
Y ’. Show that∼ is an equivalence relation (called topological equivalence).
3.23 Tell, justifying the answer, whether the group of homeomorphisms of the inter-
val[0, 1] is Abelian.
3.24 (♥) Let X, Y be topological spaces and B a topological basis of X. Prove that
f: X → Y is open if and only if f (A) is open in Y for every A ∈ B.
3.25 Let f: X → Y be a continuous mapping and A ⊂ X a dense subset. Prove
f(A) is dense in f (X).
3.26 Let f: X → Y be an open mapping and D ⊂ Y a dense subset. Prove f−1(D)
is dense in X .
3.27 Consider the following commutative diagram of continuous maps:
X f g // Z Y h ??~ ~ ~ ~ ~ ~ ~
Show that h is closed provided f is onto and g closed.
3.28 Let X be a topological space. A map f: X → R is upper semi-continuous
if f−1(] − ∞, a[) is open in X for any a ∈ R. Prove f : X → R is upper semi- continuous if and only if, for any x ∈ X and any ε > 0, there is a neighbourhood U of x such that f(y) < f (x) + ε for every y ∈ U.
3.29 (♥) Find an example of a map f : [0, 1] → [0, 1] that is continuous only at
the point 0.
3.30 (♥) Let A ⊂ R be a countable subset. Find a map f : R → [0, 1] that is
continuous at every point inR − A, and only there.
2Rigorously speaking, the notion of a category is quite different from that of a set: while we wait
3.31 (K, ♥) Let X be a topological space and f : X → R any map. Prove the subset
of points of X at which f is continuous is the intersection of a countable family of open sets.
Remark 3.32 It can be proved (Exercise6.25) that the set of rational numbers cannot be written as countable intersection of open sets inR in any way. A baffling con- sequence of Exercises3.30and3.31is that there do exist maps that are continuous only at irrational points, while there’s no continuous mapping onQ only.