Turn on thread page Beta
    • Thread Starter
    Offline

    2
    ReputationRep:
    I need to show that the contractible space X is path connected.

    The proof we were shown goes like this:

    X is contractible implies there exists a map h:X \times I \rightarrow X such that h(x,0)=x,h(x,1)=p for all x \in X and some p \in X. Now take arbitraryx_0,y_0 \in X and consider the map q from x_0 to y_0 given by:
    q:[0,1] \rightarrow X; t \mapsto \begin{cases} h(x_o,2t), \quad \text{ for } t \leq \frac{1}{2} \\ h(y_0,2-2t), \quad \text{ for } t \geq \frac{1}{2} \end{cases}

    i missed the lecture on this and have a few concerns. namely, surely we have to verify certain properties of this path e.g. continuity (otherwise X won't be path connected).

    also, at the begining, they say that X being contractible implies the existance of the map h but i can't find this in the notes. we were told that X being contractible means it's homotopy equivalent to the space \{ 0 \} with on point. are these definitions equivalent? if so, how?

    thanks.
    • Wiki Support Team
    Offline

    14
    ReputationRep:
    Wiki Support Team
    X is contractible if that map h exists and is continuous (some lecturers will just say "a map" to mean "a continuous map") - that should answer you as to whether q is continuous or not.

    Suppose X is contractible, i.e. it's homotopy equivalent to {0}. Then by definition, there are (continuous) maps f: X --> {0}, g: {0} --> X, such that gf (the composition) is homotopic to the identity map on X and fg is homotopic to the identity map on {0}. Let's be more precise; f is obviously the zero function sending everything to 0, and g sends 0 to g(0), which I'll call p. So gf is the map sending every element of X to p (fg isn't relevant here - can you see why?), and by assumption that X is contractible, we know that gf is homotopic to id_X. That is, there is a homotopy h: X*I --> X such that h(x,0) = id_X(x) = x and h(x,1) = gf(x) = p.

    This is actually a really simple argument, but I've left in all the details in case you're still finding the definitions tricky - the idea of contractibility is that we can shrink X down to a point continuously (doesn't matter whether the point is called 0 or p), and both of the definitions given seem to satisfy this. I've shown that your definition implies the one in the proof. (You can try to prove the converse to show the definitions are equivalent, if you like.)
    Offline

    13
    ReputationRep:
    You know by continuity of h that q is continuous on each of A=[0, 0.5] and B=[0.5, 1]. Then if C is closed in X, it's preimage under the restriction of q to A is closed in A and hence in [0,1] as A is closed in [0,1]. Ditto with B instead of A. Hence, preimage of C under q is closed and q is continuous.
    • Thread Starter
    Offline

    2
    ReputationRep:
    what is C in your definition? the image of q?
    Online

    18
    ReputationRep:
    As I understand SsEe's post, C is an arbitrary closed set. For any function f, if the preimage of every closed set is closed, then f is cts.
    • Thread Starter
    Offline

    2
    ReputationRep:
    so my interpretation of SsEe's post is this:

    the defn of q implies q is continuous on A and B because h is a continuous function. if we take an arbitrary C in X, p^{-1}|_A(C) is closed in A since q|_A is continuous (we noted above q is continuous on A i.e. the restriciton of q to A is continuous). similarly q^{-1}|_B(C) is closed in B since q|_B is continuous.
    if the preimage of C is closed in A then it is closed in [0,1] since A is closed in [0,1] and simliarly B is closed in [0,1] so if the preimage of C is closed in B then it is closed in [0,1].
    this means that for an arbitrary closed set C in X, q^{-1}(C) will be closed in [0,1] i.e. q is a continuous function.

    is that correct?

    (N.B. how come h(x_0,1/2) doesn't equal h(y_0,1/2). doesn't this imply a discontinuity at t=1/2)

    so we have found this continuous map q:I \rightarrow X s.t. q(0)=x_0,q(1)=y_0 for arbitrary x_0,y_0 \in X. this means the space X is path connected.

    does that look like everything's been covered in full now?

    thanks.
    Offline

    13
    ReputationRep:
    (Original post by latentcorpse)
    so my interpretation of SsEe's post is this:

    the defn of q implies q is continuous on A and B because h is a continuous function. if we take an arbitrary C in X, p^{-1}|_A(C) is closed in A since q|_A is continuous (we noted above q is continuous on A i.e. the restriciton of q to A is continuous). similarly q^{-1}|_B(C) is closed in B since q|_B is continuous.
    if the preimage of C is closed in A then it is closed in [0,1] since A is closed in [0,1] and simliarly B is closed in [0,1] so if the preimage of C is closed in B then it is closed in [0,1].
    this means that for an arbitrary closed set C in X, q^{-1}(C) will be closed in [0,1] i.e. q is a continuous function.

    is that correct?
    Yep.

    (N.B. how come h(x_0,1/2) doesn't equal h(y_0,1/2). doesn't this imply a discontinuity at t=1/2)
    At t=1/2, q(t) = h(x_0, 1) = h(y_0, 1) = p. (basically, you haven't put t=1/2 into the definition of q correctly).


    so we have found this continuous map q:I \rightarrow X s.t. q(0)=x_0,q(1)-y_0 for arbitrary x_0,y_0 \in X. this means the space X is path connected.

    does that look like everything's been covered in full now?

    thanks.
    That's the definition of path connected. So yep. All done.
 
 
 

University open days

  1. University of Bradford
    University-wide Postgraduate
    Wed, 25 Jul '18
  2. University of Buckingham
    Psychology Taster Tutorial Undergraduate
    Wed, 25 Jul '18
  3. Bournemouth University
    Clearing Campus Visit Undergraduate
    Wed, 1 Aug '18
Poll
How are you feeling in the run-up to Results Day 2018?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.