Abstract Nonsense

Crushing one theorem at a time

Homotopy and the Homotopy Category (Pt. II)


Point of Post: This is a continuation of this post.

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Although not entirely obvious, the notion of being ‘homotopic’ is actually and equivalence relation on the set C(X,Y) of all continuous maps X\to Y. Indeed, it’s clear that f\simeq f for every f\in C(X,Y) since H(x,t)=f(x) for all t is a homotopy from f to itself [this is just sequence of maps that just stays stationary at f for the entire time interval [0,1]). The relation is symmetric for if f\simeq g with homotopy H(x,t) then defining H'(x,t)=H(x,1-t) is a continuous map X\times[0,1]\to Y with

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H'(x,0)=H(x,1)=g

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and

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H'(x,1)=H(x,0)=f

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(intuitively, this is just doing the deformation from f to g, but in reverse) and thus g\simeq f. Lastly, suppose that we have a third map h:Y\to Z. Moreover, suppose that f\simeq g by homotopy H and g\simeq h by homotopy F then we can define a homotopy from f to h by the map G:X\times[0,1]\to Z given by

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\displaystyle G(x,t)=\begin{cases}H(x,2t) & \mbox{if}\quad t\in[0,\tfrac{1}{2}]\\ F(x,2t-1) & \mbox{if}\quad t\in[\tfrac{1}{2},1]\end{cases}

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This map is continuous since we have the closed cover \{X\times[0,\tfrac{1}{2}],X\times[\tfrac{1}{2},1]\} of X\times[0,1], the restriction to each of these pieces is continuous (obviously), and they agree on the intersection of the pieces (the only intersection points are X\times\{\tfrac{1}{2}\} for which H(x,2(\tfrac{1}{2}))=H(x,1)=g(x)=F(x,0)=F(x,2(\tfrac{1}{2})-1))–the rest then follows from the gluing lemma.

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We may summarize the above discussion with the following:

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Theorem: Let X and Y be topological spaces. Then, \simeq is an equivalence relation on C(X,Y).

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We denote the set C(X,Y)/\simeq of all homotopy clases (i.e. equivalence classes under \simeq) by [X,Y]. A typical element of [X,Y] will be denoted [f].

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Now that we understands for two maps f and g to be homotopic we can define the notion of a homotopy equivalence between two spaces X and Y. In particular, we say that a continuous map f:X\to Y is a homotopy equivalence if there exists a continuous map g:Y\to X such that g\circ f\simeq\text{id}_X and f\circ g\simeq\text{id}_Y. We say that two spaces are homotopy equivalent or of the same homotopy type if there exists a homotopy equivalence between them. We denote the relation “X is homotopy equivalent to Y” as X\simeq Y. It’s clear that being homotopy equivalent is an equivalence relation on the ‘set’ of all topological spaces.

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We call a space X contractible if it is homotopy equivalent to a point. For example, we have shown above that the unit disc \mathbb{D}\subseteq\mathbb{C} is contractible. This represents, in some sense, the ultimate level of connectivity since the point is as connected as a space can get, and our space X is equivalent to a point in a connectedness-focused way.

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Now, while this is a very nice way to think about the contractibility of a space, there are in fact many such ways–we discuss only one here. We call a continuous map X\to Y null-homotopic if it is homotopic to a constant map X\to Y.

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Theorem: A topological space X is contractible if and only if the identity map \text{id}_X:X\to X is null-homotopic.

Proof: Suppose first that X is contractible. Then, there exists a one-point space \{p\} and a homotopy equivalence f:X\to\{p\}. We note then that f\circ f^{-1} is a map X\to X and since it factors through \{p\} we necessarily see that f\circ f^{-1} is a constant map. That said, by assumption f\circ f^{-1}\simeq\text{id}_X.

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Conversely, suppose that \text{id}_X\simeq c for some constant map c:X\to X. Consider then \text{im }c=\{p\}\subseteq X. Then, we see that c is in fact a continuous map X\to\{p\}. Note then that if we define i to be the inclusion \{p\}\subseteq X we have that c\circ i=\text{id}_{\{p\}}\simeq\text{id}_{\{p\}} and i\circ c=c\simeq\text{id}_X. Thus, we see that c is a homotopy equivalence X\to\{p\} and thus X is contractible. \blacksquare

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Remark: This is a good time to point out something to be careful of when one is working with homotopies. This is best expressed through a cautionary tale. When I was younger I was convinced for a whole hour that the sphere \mathbb{S}^2\subseteq\mathbb{R}^3 was contractible. My argument was that one could keep shrinking down the sphere smaller and smaller until it was eventually a point. In particular, one would start with the unit sphere \mathbb{S}^2, and at time t one would consider the scared sphere t\mathbb{S}^2. Eventually one would end up with the one-point space  0\mathbb{S}^2=\{(0,0,0)\}. Thus, with slightly more formality to this “argument”, I had proven that the sphere \mathbb{S}^2 was contractible. That said, from my intuition about what contractible means I knew this could not be true since \mathbb{S}^2 had a big whole taken up by it’s hollow interior! Of course, the astute reader will recognize my error immediately–the constant map sending everything to the origin is not a constant map \mathbb{S}^2\to\mathbb{S}^2–it leaves the space. The basic idea is that homotopies need to stay within your space, whereas I left my space (by shrinking it!) which is a big no-no. In fact, one will observe that what I described was (exactly the same as for why the disc \mathbb{D} is contractible) why the unit ball B_1((0,0,0))\subseteq\mathbb{R}^3 is contractible. A subtle, but very important point to notice.

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References:

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[1] Spanier, Edwin Henry. Algebraic Topology. New York: McGraw-Hill, 1966. Print.

[2] Hatcher, Allen. Algebraic Topology. Cambridge: Cambridge UP, 2002. Print.

[3] Bredon, Glen E. Topology and Geometry. New York: Springer-Verlag, 1993. Print.

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August 30, 2012 - Posted by | Algebraic Topology, Topology | , , , , ,

1 Comment »

  1. […] Homotopy and the Homotopy Category (Pt. III) Point of Post: This is a continuation of this post. […]

    Pingback by Homotopy and the Homotopy Category (Pt. III) « Abstract Nonsense | August 30, 2012 | Reply


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