2.1 Maps

2.1.1 Definitions

Let $X$ and $Y$ be sets. A map (or mapping) $f$ is a rule by which we assign $y\in Y$ for each $x\in X$ . We write

$$f:X\to Y.\text{\hspace{1em}(2.1)}$$

If $f$ is defined by some explicit formula, we may write

$$f:x\mapsto f(x)\text{\hspace{1em}(2.2)}$$

There may be more than two elements in $X$ that correspond to the same $y\in Y$ . A subset of $X$ whose elements are mapped to $y\in Y$ under $f$ is called the inverse image of $y$ , denoted by $f^{-1}(y)=\{x\in X|f(x)=y\}$ . The set $X$ is called the domain of the map while $Y$ is called the range of the map. The image of the map is $f(X)=\{y\in Y|y=f(x)\text{ for some }x\in X\}\subset Y$ . The image $f(X)$ is also denoted by $\mathrm{im}f$ . The reader should note that a map cannot be defined without specifying the domain and the range. Take $f(x)=\exp x$ , for example. If both the domain and the range are $\mathbb{R}$ , $f(x)=-1$ has no inverse

image. If, however, the domain and the range are the complex plane $\mathbb{C}$ , we find $f^{-1}(-1)=\{(2n+1)\pi \mathrm{i}|n\in Z\}$ . The domain $X$ and the range $Y$ are as important as $f$ itself in specifying a map.

Example 2.1

Let $f:\mathbb{R}\to \mathbb{R}$ be given by $f(x)=\sin x$ . We also write $f:x\mapsto \sin x$ . The domain and the range are $\mathbb{R}$ and the image $f(\mathbb{R})$ is $[-1,1]$ . The inverse image of $0$ is $f^{-1}(0)=\{n\pi |n\in \mathbb{Z}\}$ . Let us take the same function $f(x)=\sin x=({\mathrm{e}}^{\mathrm{i}x}-{\mathrm{e}}^{-\mathrm{i}x})/2\mathrm{i}$ but $f:\mathbb{C}\to \mathbb{C}$ this time. The image $f(C)$ is the whole complex plane $\mathbb{C}$ .

Definition 2.1

If a map satisfies a certain condition it bears a special name.

(a) A map $f:X\to Y$ is called injective (or one to one) if $x\ne x^{\prime }$ implies $f(x)\ne f(x^{\prime })$ .
(b) A map $f:X\to Y$ is called surjective (or onto) if for each $y\in Y$ there exists at least one element $x\in X$ such that $f(x)=y$ .
(c) A map $f:X\to Y$ is called bijective if it is both injective and surjective.

Example 2.2

A map $f:\mathbb{R}\to \mathbb{R}$ defined by $f:x\mapsto ax$ ( $a\in \mathbb{R}-\{0\}$ ) is bijective. $f:\mathbb{R}\to \mathbb{R}$ defined by $f:x\mapsto x^2$ is neither injective nor surjective. $f:\mathbb{R}\to \mathbb{R}$ given by $f:x\mapsto \exp x$ is injective but not surjective.

Exercise 2.1

A map $f:\mathbb{R}\to \mathbb{R}$ defined by $f:x\mapsto \sin x$ is neither injective nor surjective. Restrict the domain and the range to make $f$ bijective.

Example 2.3

Let $M$ be an element of the general linear group $\mathrm{GL}(n,\mathbb{R})$ whose matrix representation is given by $n\times n$ matrices with non-vanishing determinant. Then $M:{\mathbb{R}}^n\to {\mathbb{R}}^n$ , $x\mapsto Mx$ is bijective. If $\det M=0$ , it is neither injective nor surjective.

A constant map $c:X\to Y$ is defined by $c(x)=y_0$ where $y_0$ is a fixed element in $Y$ and $x$ is an arbitrary element in $X$ . Given a map $f:X\to Y$ , we may think of its restriction to $A\subset X$ , which is denoted as $f{|}_A:A\to Y$ . Given two maps $f:X\to Y$ and $g:Y\to Z$ , the composite map of $f$ and $g$ is a map $g\circ f:X\to Z$ defined by $g\circ f(x)=g(f(x))$ . A diagram of maps is called commutative if any composite maps between a pair of sets do not depend on how they are composed. For example, in figure 2.1, $f\circ g=h\circ j$ and $f\circ g=k$ etc.

Figure 2.1. A commutative diagram of maps.

Exercise 2.2

Let $f:\mathbb{R}\to \mathbb{R}$ be defined by $f:x\to x^2$ and $g:\mathbb{R}\to \mathbb{R}$ by $g:x\to \exp x$ . What are $g\circ f:\mathbb{R}\to \mathbb{R}$ and $f\circ g:\mathbb{R}\to \mathbb{R}$ ?

If $A\subset X$ , an inclusion map $i:A\to X$ is defined by $i(a)=a$ for any $a\in A$ . An inclusion map is often written as $i:A\hookrightarrow X$ . The identity map ${\mathrm{id}}_X:X\to X$ is a special case of an inclusion map, for which $A=X$ . If $f:X\to Y$ defined by $f:x\mapsto f(x)$ is bijective, there exists an inverse map $f^{-1}:Y\to X$ , such that $f^{-1}:f(x)\to x$ , which is also bijective. The maps $f$

and $f^{-1}$ satisfy $f\circ f^{-1}={\mathrm{id}}_Y$ and $f^{-1}\circ f={\mathrm{id}}_X$ . Conversely, if $f:X\to Y$ and $g:Y\to X$ satisfy $f\circ g={\mathrm{id}}_Y$ and $g\circ f={\mathrm{id}}_X$ , then $f$ and $g$ are bijections. This can be proved from the following exercise.

Exercise 2.3

Show that if $f:X\to Y$ and $g:Y\to X$ satisfy $g\circ f={\mathrm{id}}_X$ , $f$ is injective and $g$ is surjective. If this is applied to $f\circ g={\mathrm{id}}_Y$ as well, we obtain the previous result.

Example 2.4

Let $f:\mathbb{R}\to (0,\infty )$ be a bijection defined by $f:x\mapsto \exp x$ . Then the inverse map $f^{-1}:(0,\infty )\to \mathbb{R}$ is $f^{-1}:x\mapsto \ln x$ . Let $g:(-\pi /2,\pi /2)\to (-1,1)$ be a bijection defined by $g:x\to \sin x$ . The inverse map is $g^{-1}:x\mapsto {\sin }^{-1}x$ .

Exercise 2.4

The $n$ -dimensional Euclidean group $E^n$ is made of an $n$ -dimensional translation $a:x\to x+a$ ( $x,a\in {\mathbb{R}}^n$ ) and an $\mathrm{O}(n)$ rotation $R:x\to Rx$ , $R\in \mathrm{O}(n)$ . A general element $(R,a)$ of $E^n$ acts on $x$ by $(R,a):x\mapsto Rx+a$ . The product is defined by $(R_2,a_2)\times (R_1,a_1):x\mapsto R_2(R_{1}x+a_1)+a_2$ , that is, $(R_2,a_2)\circ (R_1,a_1)=(R_2{R}_1,R_2{a}_1+a_2)$ . Show that the maps $a$ , $R$ and $(R,a)$ are bijections. Find their inverse maps.

Suppose certain algebraic structures (product or addition, say) are endowed with the sets $X$ and $Y$ . If $f:X\to Y$ preserves these algebraic structures, then $f$ is called a homomorphism. For example, let $X$ be endowed with a product. If $f$ is a homomorphism, it preserves the product, $f(ab)=f(a)f(b)$ . Note that $ab$ is defined by the product rule in $X$ , and $f(a)f(b)$ by that in $Y$ . If a homomorphism $f$ is bijective, $f$ is called an isomorphism and $X$ is said to be isomorphic to $Y$ , denoted $x\cong y$ .

2.1.2 Equivalence relation and equivalence class

Some of the most important concepts in mathematics are equivalence relations and equivalence classes. Although these subjects are not directly related to maps, it is appropriate to define them at this point before we proceed further. A relation $R$ defined in a set $X$ is a subset of $X^2$ . If a point $(a,b)\in X^2$ is in $R$ , we may write $aRb$ . For example, the relation $>$ is a subset of ${\mathbb{R}}^2$ . If $(a,b)\in >$ , then $a>b$ .

Definition 2.2

An equivalence relation $\sim$ is a relation which satisfies the following requirements:

(i) $a\sim a$ (reflective).
(ii) If $a\sim b$ , then $b\sim a$ (symmetric).
(iii) If $a\sim b$ and $b\sim c$ , then $a\sim c$ (transitive).

Exercise 2.5

If an integer is divided by 2, the remainder is either 0 or 1. If two integers $n$ and $m$ yield the same remainder, we write $m\sim n$ . Show that $\sim$ is an equivalence relation in $\mathbb{Z}$ .

Given a set $X$ and an equivalence relation $\sim$ , we have a partition of $X$ into mutually disjoint subsets called equivalence classes. A class $[a]$ is made of all the elements $x$ in $X$ such that $x\sim a$ ,

$$[a]=\{x\in X|x\sim a\}\text{\hspace{1em}(2.3)}$$

$[a]$ cannot be empty since $a\sim a$ . We now prove that if $[a]\cap [b]\ne \varnothing$ then $[a]=[b]$ . First note that $a\sim b$ . (Since $[a]\cap [b]\ne \varnothing$ there is at least one element in $[a]\cap [b]$ that satisfies $c\sim a$ and $c\sim b$ . From the transitivity, we have $a\sim b$ .) Next we show that $[a]\subset [b]$ . Take an arbitrary element $a^{\prime }$ in $[a]$ ; $a^{\prime }\sim a$ . Then $a\sim b$ implies $b\sim a^{\prime }$ , that is $a^{\prime }\in [b]$ . Thus, we have $[a]\subset [b]$ . Similarly, $[a]\supset [b]$ can be shown and it follows that $[a]=[b]$ . Hence, two classes $[a]$ and $[b]$ satisfy either $[a]=[b]$ or $[a]\cap [b]=\varnothing$ . In this way a set $X$ is decomposed into mutually disjoint equivalence classes. The set of all classes is called the quotient space, denoted by $X/\sim$ . The element $a$ (or any element in $[a]$ ) is called the representative of a class $[a]$ . In exercise 2.5, the equivalence relation $\sim$ divides integers into two classes, even integers and odd integers. We may choose the representative of the even class to be 0, and that of the odd class to be 1. We write this quotient space $\mathbb{Z}/\sim$ . $\mathbb{Z}/\sim$ is isomorphic to ${\mathbb{Z}}_2$ , the cyclic group of order 2, whose algebra is defined by $0+0=0$ , $0+1=1+0=1$ and $1+1=0$ . If all integers are divided into equivalence classes according to the remainder of division by $n$ , the quotient space is isomorphic to ${\mathbb{Z}}_n$ , the cyclic group of order $n$ .

Let $X$ be a space in our usual sense. (To be more precise, we need the notion of topological space, which will be defined in section 2.3. For the time being we depend on our intuitive notion of ‘space’.) Then quotient spaces may be realized as geometrical figures. For example, let $x$ and $y$ be two points in $\mathbb{R}$ .

(a)

(b)

Figure 2.2. In (a) all the points $x+2n\pi$ , $n\in \mathbb{Z}$ are in the same equivalence class $[x]$ . We may take $x\in [0,2\pi )$ as a representative of $[x]$ . (b) The quotient space $\mathbb{R}/\sim$ is the circle $S^1$ .

Introduce a relation $\sim$ by: $x\sim y$ if there exists $n\in \mathbb{Z}$ such that $y=x+2\pi n$ . It is easily shown that $\sim$ is an equivalence relation. The class $[x]$ is the set $\{\dots ,x-2\pi ,x,x+2\pi ,\dots \}$ . A number $x\in [0,2\pi )$ serves as a representative of an equivalence class $[x]$ , see figure $2.2(a)$ . Note that $0$ and $2\pi$ are different points in $\mathbb{R}$ but, according to the equivalence relation, these points are looked upon as the same element in $\mathbb{R}/\sim$ . We arrive at the conclusion that the quotient space $\mathbb{R}/\sim$ is the circle $S^1=\{{\mathrm{e}}^{\mathrm{i}\theta }|0\le \theta <2\pi \}$ ; see figure $2.2(b)$ . Note that a point $\epsilon$ is close to a point $2\pi -\epsilon$ for infinitesimal $\epsilon$ . Certainly this is the case for $S^1$ , where an angle $\epsilon$ is close to an angle $2\pi -\epsilon$ , but not the case for $\mathbb{R}$ . The concept of closeness of points is one of the main ingredients of topology.

Example 2.5

(a) Let $X$ be a square disc $\{(x,y)\in {\mathbb{R}}^2||x|\ge 1,|y|\ge 1\}$ . If we identify the points on a pair of facing edges, $(-1,y)\sim (1,y)$ , for example, we obtain the cylinder, see figure 2.3(a). If we identify the points $(-1,-y)\sim (1,y)$ , we find the Möbius strip, see figure 2.3(b). [Remarks: If readers are not familiar with the Möbius strip, they may take a strip of paper and glue up its ends after a $\pi$ -twist. Because of the twist, one side of the strip has been joined to the other side, making the surface single sided. The Möbius strip is an example of a non-orientable surface, while the cylinder has definite sides and is said to be orientable. Orientability will be discussed in terms of differential forms in section 5.5.]

Figure 2.3. (a) The edges $|x|=1$ are identified in the direction of the arrows to form a cylinder. (b) If the edges are identified in the opposite direction, we have a Möbius strip.

(b) Let $(x_1,y_1)$ and $(x_2,y_2)$ be two points in ${\mathbb{R}}^2$ and introduce an equivalence relation $\sim$ by: $(x_1,y_1)\sim (x_2,y_2)$ if $x_2=x_1+2\pi n_x$ and $y_2=y_1+2\pi n_y$ , $n_x,n_y\in \mathbb{Z}$ . Then $\sim$ is an equivalence relation. The quotient space ${\mathbb{R}}_2/\sim$ is the torus $T^2$ (the surface of a doughnut), see figure 2.4(a). Alternatively, $T^2$ is

Figure 2.4. If all the points $(x+2\pi n_x,y+2\pi n_y),n_x,n_y\in \mathbb{Z}$ are identified as in (a), the quotient space is taken to be the shaded area whose edges are identified as in (b). This resulting quotient space is the torus $T^2$ .

represented by a rectangle whose edges are identified as in figure 2.4(b).

(c) What if we identify the edges of a rectangle in other ways? Figure 2.5 gives possible identifications. The spaces obtained by these identifications are

(a)

(b)

Figure 2.5. The Klein bottle $(a)$ and the projective plane $(b)$ .

called the Klein bottle, figure 2.5(a), and the projective plane, figure 2.5(b), neither of which can be realized (or embedded) in the Euclidean space ${\mathbb{R}}^3$ without intersecting with itself. They are known to be non-orientable.

The projective plane, which we denote $R{P}^2$ , is visualized as follows. Let us consider a unit vector $n$ and identify $n$ with $-n$ , see figure 2.6. This identification takes place when we describe a rod with no head or tail, for example. We are tempted to assign a point on $S^2$ to specify the ‘vector’ $n$ . This works except for one point. Two antipodal points $n=(\theta ,\varphi )$ and $-n=(\pi -\theta ,\pi +\varphi )$ represent the same state. Then we may take a northern hemisphere as the coset space $S^2/\sim$ since only a half of $S^2$ is required. However, the coset space is not just an ordinary hemisphere since the antipodal points on the equator are identified. By continuous deformation of this hemisphere into a square, we obtain the square in figure 2.5(b).

Figure 2.6. If $n$ has no head or tail, one cannot distinguish $n$ from $-n$ and they must be identified. One obtains the projective plane $R{P}^2$ by this identification $n\sim -n$ ; $R{P}^2\simeq S^2/\sim$ . It suffices to take a hemisphere to describe the coset space. Note, however, that the antipodal points on the equator are identified.

(d) Let us identify pairs of edges of the octagon shown in figure 2.7(a). The quotient space is the torus with two handles, denoted by ${\Sigma }_2$ , see figure 2.7(b). ${\Sigma }_g$ , the torus with $g$ handles, can be obtained by a similar identification, see problem 2.1. The integer $g$ is called the genus of the torus.
(e) Let $D^2=\{(x,y)\in {\mathbb{R}}^2|x^2+y^2\le 1\}$ be a closed disc. Identify the points on the boundary $\{(x,y)\in {\mathbb{R}}^2|x^2+y^2=1\}$ ; $(x_1,y_1)\sim (x_2,y_2)$ if $x_1^2+y_1^2=x_2^2+y_2^2=1$ . Then we obtain the sphere $S^2$ as the quotient space $D^2/\sim$ , also written as $D^2/S^1$ , see figure 2.8. If we take an $n$ -dimensional disc $D^n=\{(x_0,\dots ,x^n)\in {\mathbb{R}}^{n+1}|(x_0{)}^2+\cdots +(x^n{)}^2\le 1\}$ and identify the points on the surface $S^{n-1}$ , we obtain the $n$ -sphere $S^n$ , namely $D^n/S^{n-1}=S^n$ .

(a)

Figure 2.7. If the edges of $(a)$ are identified a torus with two holes (genus two) is obtained.

Figure 2.8. A disc $D^2$ whose boundary $S^1$ is identified is the sphere $S^2$ .

Exercise 2.6

Let $\mathrm{H}$ be the upper-half complex plane $\{\tau \in \mathbb{C}|\mathrm{Im}\tau \ge 0\}$ . Define a

group

$$\mathrm{SL}(2,\mathbb{Z})\equiv \left\{\left(\begin{array}{ll}a& b\\ c& d\end{array}\right)|a,b,c,d\in \mathbb{Z},ad-bc=1\right\}.\text{\hspace{1em}(2.4)}$$

Introduce a relation $\sim$ , for $\tau ,{\tau }^{\prime }\in \mathrm{H}$ , by $\tau \sim {\tau }^{\prime }$ if there exists a matrix

$$A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in \mathrm{SL}(2,\mathbb{Z})$$

such that

$${\tau }^{\prime }=(a\tau +b)/(c\tau +d).\text{\hspace{1em}(2.5)}$$

Show that this is an equivalence relation. (The quotient space $\mathrm{H}/\mathrm{SL}(2,\mathbb{Z})$ is shown in figure 8.3.)

Example 2.6

Let $G$ be a group and $H$ a subgroup of $G$ . Let $g,g^{\prime }\in G$ and introduce an equivalence relation $\sim$ by $g\sim g^{\prime }$ if there exists $h\in H$ such that $g^{\prime }=gh$ . We denote the equivalence class $[g]=\{gh|h\in H\}$ by $gH$ . The class $gH$ is called a (left) coset. $gH$ satisfies either $gH\cap g^{\prime }H=\varnothing$ or $gH=g^{\prime }H$ . The quotient space is denoted by $G/H$ . In general $G/H$ is not a group unless $H$ is a normal subgroup of $G$ , that is, $gh{g}^{-1}\in H$ for any $g\in G$ and $h\in H$ . If $H$ is a normal subgroup of $G$ , $G/H$ is called the quotient group, whose group operation is given by $[g]\ast [g^{\prime }]=[g{g}^{\prime }]$ , where $\ast$ is the product in $G/H$ . Take $ghϵ[g]$ and $g^{\prime }{h}^{\prime }ϵ[g^{\prime }]$ . Then there exists $h^{\prime \prime }ϵH$ such that $h{g}^{\prime }=g^{\prime }{h}^{\prime \prime }$ and hence $gh{g}^{\prime }{h}^{\prime }=g{g}^{\prime }{h}^{\prime \prime }{h}^{\prime }ϵ[g{g}^{\prime }]$ . The unit element of $G/H$ is the equivalence class $[e]$ and the inverse element of $[g]$ is $[g^{-1}]$ .

Exercise 2.7

Let $G$ be a group. Two elements $a,b\in G$ are said to be conjugate to each other, denoted by $a\simeq b$ , if there exists $g\in G$ such that $b=ga{g}^{-1}$ . Show that $\simeq$ is an equivalence relation. The equivalence class $[a]=\{ga{g}^{-1}|g\in G\}$ is called the conjugacy class.

来源与版本

正文：英文 · OCR 机器稿 · 待校对

核对状态：OCR 机器稿 · 待校对

原书 PDF

来源版本：second-edition-local

来源 PDF SHA-256：fcd24c3371d092a8c699ccf5cf0765462924b6c48412804231076e8084b1940d

OCR 来源 SHA-256：0b6324f01172a3548cd34382e82fd8bfae574602b6b2cfaa357ba96bc2c35478

OCR 产物 SHA-256：57a64667da4d3fccda699a9230c2383dc1c9304631645626e3b28e3aa18dc71d