M40: Exercise sheet 5

M40: Exercise sheet 5
1. Let X be a topological space. Recall that the path components of X are the equivalence
classes in X under the equivalence relation where x and y are related if and only if there
is a path from x to y. (For X locally path-connected, these are just the connected
components of X.) Denote the set of path components of X by π0 (X). Now let
φ : X → Y be a continuous map.
(a) Define a map φ∗ : π0 (X) → π0 (Y ). (In particular, show it is well-defined).
(b) Show that if φ ' ψ then φ∗ = ψ∗ .
(c) Show that if f : X → Y and g : Y → Z then (g ◦ f )∗ = g∗ ◦ f∗ while (idX )∗ =
idπ0 (X) .
(Thus we get another example of a functor : this time from topological spaces to
sets.)
(d) Finally prove that if φ is a homotopy equivalence then φ∗ is a bijection.
2. Let X have the property that every continuous map S 1 → X is homotopic to a
constant. Prove that for any x ∈ X, π1 (X, x) = {1}. Hint: View a loop as a map
S 1 → X (but be careful about basepoints!).
3. (Lebesgue Covering Lemma) Let (X, d) be a compact metric space and {Uα } an open
cover of X. Show that there is a number δ > 0 such that any subset of X of diameter
less than δ is completely contained in some Uα .
4. Let X, Y be topological spaces with Y Hausdorff. Show that C (X, Y ) (with the
compact-open topology) is Hausdorff also.
5. Let X, Y be topological spaces and, as usual, give C (X, Y ) the compact-open topology.
(a) (Easy) For x ∈ X, define evx : C (X, Y ) → Y by
evx (φ) = φ(x).
Show that evx is continuous.
(b) (A little harder) Now suppose that X has the property that any neighbourhood
of any x ∈ X contains a compact neighbourhood of x (we say that X is strongly
locally compact 1 ). Define ev : X × C (X, Y ) → Y by
ev(x, φ) = φ(x).
Show that ev is continuous (here, of course, we give X × C (X, Y ) the product
topology).
November 6, 2014
1
Observe that any interval in R has this property.
Home page: http://go.bath.ac.uk/ma40040
M40: Exercise sheet 5—Solutions
1.
(a) Let Cx denote the path component of x and define φ∗ : π0 (X) → π0 (Y ) by φ∗ Cx = Cφ(x) .
To see that this is well defined, suppose Cx1 = Cx2 . This means there is a path, γ say,
from x1 to x2 . But then φ ◦ γ is a path from φ(x1 ) to φ(x2 ) so that Cφ(x1 ) = Cφ(x2 ) .
(b) Let F be a homotopy from φ to ψ and let x ∈ X. Then t 7→ F (x, t) is a path from φ(x)
to ψ(x) so that Cφ(x) = Cψ(x) , i.e., φ∗ (Cx ) = ψ∗ (Cx ) as required.
(c) Fix x ∈ X. Then g∗ ◦ f∗ (Cx ) = g∗ (Cf (x) ) = Cg(f (x)) = (g ◦ f )∗ (Cx ). Moreover,
(idX )∗ (Cx ) = Cx = idπ0 (X) (Cx ).
(d) Finally, if f : X → Y is a homotopy equivalence with homotopy inverse g then g ◦f ' idX
so that g∗ ◦ f∗ = (g ◦ f )∗ = (idX )∗ = idπ0 (X) and, similarly, f∗ ◦ g∗ = idπ0 (Y ) . Thus f∗ is
a bijection with inverse g∗ .
2.
We may identify loops in X with maps of S 1 into X as follows: if α is a loop in X, define
α
ˆ : S 1 → X by α
ˆ (e2πit ) = α(t). It is easy to check that α 7→ α
ˆ is a bijection from loops based
ˆ
at some x ∈ X to continuous maps (S 1 , 1) → (X, x) and that α ' β if and only if α
ˆ ' β.
With this in hand, suppose now that α is a loop at x and that α
ˆ is homotopic to a constant
map with value y, say. Then α ' γy via some homotopy F with F (0, s) = F (1, s) for each
s ∈ I. Let β be the path from x to y given by β(s) = F (0, s) then, from lemma (2.13) of the
lecture notes, we have
α ∼ β · γy · β ∼ γx ,
as required.
3.
Let x ∈ X. Then x ∈ Uα for some α(x) and there is δx > 0 so that the 2δx ball B2δx (x)
about x is contained in Uα . Then {Bδx (x)}x∈X is an open cover for X and so there is a finite
sub-cover {Bδxi (xi )}1≤i≤n .
Now set δ = min1≤i≤n δi > 0 and suppose that A ⊂ X has diameter less than δ. Fix an a ∈ A.
Then a ∈ Bδxi (xi ) for some i whence the triangle inequality gives A ⊂ B2δxi (xi ) which last
lies completely in some Uα as required.
4.
Suppose that f 6= g ∈ C (X, Y ). Thus, for some x ∈ X, f (x) 6= g(x). Since Y is Hausdorff,
there are disjoint open sets G1 and G2 in Y with f (x) ∈ G1 and g(x) ∈ G2 .
Contemplate now the subbasic open sets C{x},Gi , i = 1, 2, of C (X, Y ). We have f ∈ C{x},G1 ,
g ∈ C{x},G2 and
C{x},G1 ∩ C{x},G2 = C{x},G1 ∩G2 = ∅.
Thus C (X, Y ) is Hausdorff.
5.
(i) Let G be open in Y . Then
ev−1
x (G) = {f : f (x) ∈ G} = C{x},G
is a subbasic open set so that evx is indeed continuous.
(ii) Again let Y be open in G and contemplate
ev−1 (G) = {(x, f ) : f (x) ∈ G}.
Fix (x, f ) ∈ ev−1 (G). Then, since f is continuous, f −1 (G) is an open neighbourhood of x and
so contains a compact neighbourhood K of x since X is strongly locally compact. So we have
an open subset U of X with x ∈ U ⊂ K and we can form the basic open set U × CK,G of
X × C (X, Y ).
By construction, (x, f ) ∈ U × CK,G while, for any (y, g) ∈ U × CK,G , we have y ∈ K so that
g(y) ∈ G so that
(x, f ) ∈ U × CK,G ⊂ ev−1 (G).
It follows at once that ev−1 (G) is open in X × C (X, Y ).

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