Simplicial Structures in Topology

22 I Fundamental Concepts
As for the Example (I.1.7), we define the function f : D 2 ≡ → S 2 such that
⎛
�
F([x,y]) = ⎝x,2 |y|− 1�
1 − x
2
2
�
� � �
�
,s(y) �1 − x2 �
+ 4 |y|− 1�
1 − x
2
2
� �
2 ⎞
⎠
for every equivalence class [x,y] ∈ D 2 ≡,wheres(y) equals +1 ify > 0and−1 if
y < 0. Once more, f is a bijective map from a compact space to a Hausdorff space
and is, therefore, a homeomorphism.
I.1.4 Function Spaces
Let X and Y be two topological spaces and let M(X,Y ) be the set of all maps from
X to Y . For every compact space K ⊂ X andeveryopensetU ⊂ Y ,wedefinethe
set WK,U of all maps f ∈ M(X,Y ) for which f (K) ⊂ U. Theset
C = {WK,U | K ⊂ X compact and U ⊂ Y open}
is a sub-basis for a topology of M(X,Y ) known as compact-open topology.
Let the topological spaces X, Y, Z and a map f : X × Z → Y be given. For every
z0 ∈ Z, the function f (−,z0): X → Y is continuous (in fact, it is the composite of the
map f and the inclusion X ×{z0}→X × Z). Therefore, we may define a function
ˆf : Z → M(X,Y ) by requiring that ˆf (z)(x) =f (x,z) for every (z,x) ∈ Z × X. The
function ˆf is the adjoint of f .
(I.1.37) Theorem. Let M(X,Y ) be the function space of all maps from X to Y with
the compact-open topology. For every map f : X × Z → Y, its adjoint function
ˆf : Z → M(X,Y ) is continuous.
Proof. Let z0 ∈ Z and let WK,U be an element of the sub-basis for the compact-open
topology on M(X,Y ) such that ˆf (z0) ∈ WK,U, thatistosay, f (x,z0) ∈ U, forevery
x ∈ K. Then
K ×{z0}⊂ f −1 (U) ⊂ X × Z
where f −1 (U) is open; hence, f −1 (U) ∩ (K × Z) is open and contains K ×{z0}.
Since K is compact, there exists an open set W ⊂ Z such that z0 ∈ W and K ×W ⊂
f −1 (U). Therefore,
ˆf (W ) ⊂ WK,U
and so ˆf is continuous. �
Conversely, given a function ˆf : Z → M(X,Y ), wedefineafunction f : X×
Z → Y by the following condition: f (x,z) = ˆf (z)(x) for every (x,z) ∈ X × Z; in
this case too, we say that f is adjoint of ˆf . The fact that ˆf : Z → M(M,Y ) is continuous
does not necessarily imply that f : X × Z → Y is continuous; for that effect,
another condition on the space X is required. But first, let us prove the following: