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# 数学代写|拓扑学代写TOPOLOGY代考|MATH625 Homotopies of Maps and Spaces

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## 数学代写|拓扑学代写TOPOLOGY代考|Homotopies of Maps and Spaces

In the last chapter, we discussed homotopies of maps between $[0,1]$ and a topological space $X$. We can generalize this to maps between two arbitrary topological spaces $X$ and $Y$. We say that two maps $f, g: X \rightarrow Y$ are homotopic if we can continuously deform one into the other. We can express this notion more formally, in a similar manner to how we defined homotopies of maps between $[0,1]$ and $X$ :

Definition $9.1$ Suppose $X$ and $Y$ are two topological spaces, and $f, g: X \rightarrow Y$ are two continuous maps. Then a homotopy between $f$ and $g$ is a continuous map $H:[0,1] \times X \rightarrow Y$ satisfying the following properties:

$H(0, x)=f(x)$ for all $x \in X$,

$H(1, x)=g(x)$ for all $x \in X$.
If there is a homotopy between $f$ and $g$, then we say that $f$ and $g$ are homotopic. We write $f \sim g$ when $f$ and $g$ are homotopic.

Example Let $X$ be the interval $[0,1]$, and let $Y$ be the single point 0 . Then $X$ and $Y$ are homotopy equivalent. To see this, we need to define maps $f: X \rightarrow Y$ and $g: Y \rightarrow X$. We define $f(x)=0$ for all $x \in X$, and $g(0)=0$ (for the only point 0 in $Y$ ). Then $(g \circ f)(x)=0$ for all $x \in X$. To see that this is homotopic to the identity map $h(x)=x$, we need to construct a homotopy $H:[0,1] \times X \rightarrow$ $X$ between them. Our homotopy will be defined by $H(s, x)=s x$. Then we have $H(0, x)=0=(g \circ f)(x)$, and $H(1, x)=x=h(x)$. So this is a homotopy between $(g \circ f)(x)$ and the identity function on $X$.

Now we have to show that $f \circ g$ is homotopic to the identity function on $Y$. But this is easier, because both functions are the same function that sends the only point in $Y$ to itself. The homotopy $J$ between them is defined by $J(s, x)=0$.

## 数学代写|拓扑学代写TOPOLOGY代考|Computing the Fundamental Group of a Circle

So far, it is not yet clear whether the fundamental group is an interesting invariantthat is, does it ever distinguish spaces? Are there any spaces at all with nontrivial fundamental group? In case the name didn’t give it away, here’s a spoiler: yes! We will show that the circle has nontrivial fundamental group.

Before we do this, let us see intuitively why we ought to believe that the circle has nontrivial fundamental group. Suppose our circle is the set $\mathbb{S}^1=\left{(x, y): x^2+y^2=\right.$ 1} $\subset \mathbb{R}^2$. Let us pick as our basepoint the point $p=(1,0)$. Let us consider the loop $\alpha$ on the circle; $\alpha$ is a map $\alpha:[0,1] \rightarrow \mathbb{S}^1$ so that $\alpha(0)=\alpha(1)=p$, and we will choose it to be the loop $\alpha(t)=(\cos 2 \pi t, \sin 2 \pi t)$, so it is a loop of constant speed that goes around the circle once in the counterclockwise direction.

This loop appears not to be homotopic to the trivial loop: it seems that this loop goes around once, and the trivial loop goes around 0 times. But how can we prove that, by doing some clever homotopy, we can’t shrink it down to a point?

There are several ways of proving this, and the different techniques highlight different properties of fundamental groups. In this section, we’ll see a way to do it using a first example of covering spaces, while in the next chapter we’ll see a different proof. We won’t talk more about covering spaces in general in this book, but the procedure we employ here to compute fundamental groups is very general and can be used to compute the fundamental group of any reasonably nice space.
The outline of the proof is the following: We want to start with a loop on the circle, lift it up to some other space, and see what the lifted version of the loop looks like.

## 数学代写|拓扑学代写TOPOLOGY代考|Homotopies of Maps and Spaces

$H(0, x)=f(x)$ 对全部 $x \in X ，$
$H(1, x)=g(x)$ 对全部 $x \in X$.

$H:[0,1] \times X \rightarrow X$ 它们之间。我们的同伦定义为 $H(s, x)=s x$. 然后我们有 $H(0, x)=0=(g \circ f)(x)$ 和 $H(1, x)=x=h(x)$. 所以这是之间的同伦 $(g \circ f)(x)$ 和身份函数 $X$.

## MATLAB代写

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