# 数学代写|现代代数代写Modern Algebra代考|Properties of Composite Mappings (Optional)

#### Doug I. Jones

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## 数学代写|现代代数代写Modern Algebra代考|Properties of Composite Mappings (Optional)

In many cases, we will be dealing with mappings of a set into itself; that is, the domain and codomain of the mappings are the same. In these cases, the mappings $f \circ g$ and $g \circ f$ are both defined, and the question of whether $f \circ g$ and $g \circ f$ are equal arises. That is, is mapping composition commutative when the domain and codomain are equal? The following example shows that the answer is no.

Example 1 Let $\mathbf{Z}$ be the set of all integers, and let the mappings $f: \mathbf{Z} \rightarrow \mathbf{Z}$ and $g: \mathbf{Z} \rightarrow \mathbf{Z}$ be defined for each $n \in \mathbf{Z}$ by
\begin{aligned} & f(n)=2 n \ & g(n)= \begin{cases}\frac{n}{2} & \text { if } n \text { is even } \ 4 & \text { if } n \text { is odd. }\end{cases} \end{aligned}

In this case, the composition mappings $f \circ g$ and $g \circ f$ are both defined. We have, on the one hand,
\begin{aligned} (g \circ f)(n) & =g(f(n)) \ & =g(2 n) \ & =n, \end{aligned}
so $(g \circ f)(n)=n$ for all $n \in \mathbf{Z}$. On the other hand,
\begin{aligned} (f \circ g)(n) & =f(g(n)) \ & = \begin{cases}f\left(\frac{n}{2}\right)=n & \text { if } n \text { is even } \ f(4)=8 & \text { if } n \text { is odd, }\end{cases} \end{aligned}
so $f \circ g \neq g \circ f$. Thus mapping composition is not commutative.

## 数学代写|现代代数代写Modern Algebra代考|Binary Operations

We are familiar with the operations of addition, subtraction, and multiplication on real numbers. These are examples of binary operations. When we speak of a binary operation on a set, we have in mind a process that combines two elements of the set to produce a third element of the set. This third element, the result of the operation on the first two, must be unique. That is, there must be one and only one result from the combination. Also, it must always be possible to combine the two elements, no matter which two are chosen. This discussion is admittedly a bit vague, in that the terms process and combine are somewhat indefinite. To eliminate this vagueness, we make the following formal definition.

It is conventional in mathematics to assume that when a formal definition is made, it is automatically biconditional. That is, it is understood to be an “if and only if” statement, without this being written out explicitly. In Definition 1.18 , for example, it is understood as part of the definition that $f$ is a binary operation on the nonempty set $A$ if and only if $f$ is a mapping from $A \times A$ to $A$. Throughout the remainder of this book, we will adhere to this convention when we make definitions.

We now have a precise definition of the term binary operation, but some of the feel for the concept may have been lost. However, the definition gives us what we want. Suppose $f$ is a mapping from $A \times A$ to $A$. Then $f(x, y)$ is defined for every ordered pair $(x, y)$ of elements of $A$, and the image $f(x, y)$ is unique. In other words, we can combine any two elements $x$ and $y$ of $A$ to obtain a unique third element of $A$ by finding the value $f(x, y)$. The result of performing the binary operation on $x$ and $y$ is $f(x, y)$, and the only thing unfamiliar about this is the notation for the result. We are accustomed to indicating results of binary operations by symbols such as $x+y$ and $x-y$. We can use a similar notation and write $x * y$ in place of $f(x, y)$. Thus $x * y$ represents the result of an arbitrary binary operation $*$ on $A$, just as $f(x, y)$ represents the value of an arbitrary mapping from $A \times A$ to $A$.

# 现代代数代考

## 数学代写|现代代数代写Modern Algebra代考|Properties of Composite Mappings (Optional)

\begin{aligned} & f(n)=2 n \ & g(n)= \begin{cases}\frac{n}{2} & \text { if } n \text { is even } \ 4 & \text { if } n \text { is odd. }\end{cases} \end{aligned}

\begin{aligned} (g \circ f)(n) & =g(f(n)) \ & =g(2 n) \ & =n, \end{aligned}

\begin{aligned} (f \circ g)(n) & =f(g(n)) \ & = \begin{cases}f\left(\frac{n}{2}\right)=n & \text { if } n \text { is even } \ f(4)=8 & \text { if } n \text { is odd, }\end{cases} \end{aligned}

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