# Abstract Nonsense

## 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.