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    Let X be a space with base point x_0 \in X and let Y=(X \cup I) / \{ x_0 \sim 0 \} be the identification space obtained from the disjoint union X \cup I by identifying x_0 \sim 0. Prove X is a deformation retract of Y i.e. that the the inclusion f : X \rightarrow Y is a homotopy equivalence.

    so as far as i can tell, all i need to do is find a map g : Y \rightarrow X such that gf \simeq \mathbb{I}_Y and fg \simeq \mathbb{I}_X. is this correct?

    well clearly f : X \rightarrow Y ; x \mapsto x
    so let's take g : Y \rightarrow X ; y \mapsto x_0

    then fg: X \rightarrow X ; x \mapsto x_0 will be homotopic to the identity on X by the homotopy h : X \times I ; (x,t) \mapsto xt + (1-t)x_0

    and gf: Y \rightarrow Y ; y \mapsto x_0 will be homotopic to the identity on Y by the homotopy k : Y \times I ; (y,t) \mapsto yt + (1-t)x_0

    this seems to work to me but i'm pretty sure i'm missing some crucial point because i haven't really made use of the equivalence relation in any way whatsoever.

    thanks for any help.
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    What do you mean by "xt", and so on? As far as I can tell, your space doesn't have a multiplication defined.

    I haven't looked very closely; there may be other errors.
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    i don't understand your concern, either i'm being a moron or surely we just define x*t=xt
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    (Original post by latentcorpse)
    Let X be a space with base point x_0 \in X and let Y=(X \cup I) / \{ x_0 \sim 0 \} be the identification space obtained from the disjoint union X \cup I by identifying x_0 \sim 0. Prove X is a deformation retract of Y i.e. that the the inclusion f : X \rightarrow Y is a homotopy equivalence.

    so as far as i can tell, all i need to do is find a map g : Y \rightarrow X such that gf \simeq \mathbb{I}_Y and fg \simeq \mathbb{I}_X. is this correct?

    well clearly f : X \rightarrow Y ; x \mapsto x
    so let's take g : Y \rightarrow X ; y \mapsto x_0

    then fg: X \rightarrow X ; x \mapsto x_0 will be homotopic to the identity on X by the homotopy h : X \times I ; (x,t) \mapsto xt + (1-t)x_0

    and gf: Y \rightarrow Y ; y \mapsto x_0 will be homotopic to the identity on Y by the homotopy k : Y \times I ; (y,t) \mapsto yt + (1-t)x_0

    this seems to work to me but i'm pretty sure i'm missing some crucial point because i haven't really made use of the equivalence relation in any way whatsoever.

    thanks for any help.
    Helps if you just think about it in pictures. Y is just X with a copy of I glued on at x0. So just shrink I down. f:X->Y by f(x)=x is fine. Take g:Y->X to be g(y)=y if y is in X and x0 if y is in I.

    As you've defined it, gf:X->X (why do you compose maps backwards?) is not homotopic to the identity on X because X need not be contractible. h makes no sense because multiplication by reals isn't necessarily defined on X. And even if it was, if X wasn't connected then h wouldn't map into X. Ditto with Y.
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    (Original post by latentcorpse)
    i don't understand your concern, either i'm being a moron or surely we just define x*t=xt
    What does "xt" mean?

    To take a silly example: suppose X is the power set of all elements in the Periodic Table (with the discrete topology). What's x*t when x = {sodium, nitrogen}?

    You're assuming X is something nice like a vector space. It might not be.
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    so are you saying i can't use xt or that i'm on the right lines but i just need to define it first?
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    using the hint in post 4 though:

    i get gf: X \rightarrow X ; x \mapsto x \Rightarrow gf = \mathbb{I}_X \Rightarrow gf \simeq \mathbb{I}_X

    and also

    fg: Y \rightarrow Y ; y \mapsto \begin{cases} y \quad \text{if } y \in X \\ x_0 \quad \text{if } y \in I \end{cases}

    is this right so far? i'm not quite sure how to make a homotopy between fg and the identity on y. would the homotopy have two cases as well? something like

    h : Y \times I ; (y,t) \mapsto \begin{cases} yt + (1-t)c \quad \text{if } y \in X \\ x_0t + (1-t)c \quad \text{if } y \in I \end{cases}
    this then raises the issue of defining yt and im not sure how to do that?
    note that i have defined the identity on Y as
    \mathbb{I}_Y ; y \mapsto c
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    (Original post by latentcorpse)
    using the hint in post 4 though:

    i get gf: X \rightarrow X ; x \mapsto x \Rightarrow gf = \mathbb{I}_X \Rightarrow gf \simeq \mathbb{I}_X

    and also

    fg: Y \rightarrow Y ; y \mapsto \begin{cases} y \quad \text{if } y \in X \\ x_0 \quad \text{if } y \in I \end{cases}

    is this right so far? i'm not quite sure how to make a homotopy between fg and the identity on y. would the homotopy have two cases as well? something like

    h : Y \times I ; (y,t) \mapsto \begin{cases} yt + (1-t)c \quad \text{if } y \in X \\ x_0t + (1-t)c \quad \text{if } y \in I \end{cases}
    this then raises the issue of defining yt and im not sure how to do that?
    note that i have defined the identity on Y as
    \mathbb{I}_Y ; y \mapsto c
    fg is already the identity on X so the h doesn't need to do anything to points of X. On [0,1], you can multiply by t here. So just set h(y, t) = ty. So, h(y,0)=y on X and 0=x0 on I. h(y, 1) = y for all Y.

    Box.
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    thanks for the help but i don't understand this last statement

    h(y,0)=0*y=0 surely and 0 is in I but 0=x_0 and x_0 is in X. is this what you meant?
    i agree though that h(y,1)=y
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    h(y, 0) =

    y for y in X

    0 ~ x0 for y in I.


    So h(y, 0) = fg(y)
 
 
 
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