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    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.
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    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.)
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    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.
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    what is C in your definition? the image of q?
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    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.
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    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.
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    (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.
 
 
 
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