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Chapter 5: Problem 48

For any mapping \(f: A \rightarrow B,\) show that \(1_{B} \circ f=f=f \circ \mathbf{1}_{A}\)

Short Answer

Expert verified
For any mapping \(f: A \rightarrow B\), we have shown that \(1_{B} \circ f = f\) and \(f \circ 1_{A} = f\) by proving these two properties separately. We used the definitions of function composition and identity mappings, and demonstrated that \((1_{B} \circ f)(a) = f(a)\) for all \(a \in A\) and \((f \circ 1_{A})(a) = f(a)\) for all \(a \in A\).

Step by step solution

01

Prove that \(1_{B} \circ f = f\)

For this property, we first need to recall the definition of function composition: \((1_{B} \circ f)(a) = 1_{B}(f(a))\). Our goal is to show that \((1_{B} \circ f)(a) = f(a)\) for all \(a \in A\). Consider any \(a \in A\). We have: \((1_{B} \circ f)(a) = 1_{B}(f(a))\) By definition of the identity mapping \(1_{B}\), we get: \(1_{B}(f(a)) = f(a)\) Thus, we have \((1_{B} \circ f)(a) = f(a)\) for all \(a \in A\), which implies \(1_{B} \circ f = f\).
02

Prove that \(f \circ 1_{A} = f\)

For this property, we first need to recall the definition of function composition again: \((f \circ 1_{A})(a) = f(1_{A}(a))\). Our goal is to show that \((f \circ 1_{A})(a) = f(a)\) for all \(a \in A\). Consider any \(a \in A\). We have: \((f \circ 1_{A})(a) = f(1_{A}(a))\) By definition of the identity mapping \(1_{A}\), we get: \(f(1_{A}(a)) = f(a)\) Thus, we have \((f \circ 1_{A})(a) = f(a)\) for all \(a \in A\), which implies \(f \circ 1_{A} = f\). In conclusion, we have shown that for any mapping \(f: A \rightarrow B\), the identity mappings in the domain and codomain satisfy the properties \(1_{B} \circ f = f\) and \(f \circ 1_{A} = f\), as required.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mapping
In mathematics, a mapping, often referred to as a function, is a rule that associates each element of a set, called the domain, to a single element of another set, known as the codomain. Think of it as a machine into which we feed an input from the domain, and it spits out an output in the codomain鈥攃onsistently the same output for the same input.

For example, consider the function f that converts Celsius to Fahrenheit. If the domain is a set of temperatures in Celsius, then the codomain would be a set of temperatures in Fahrenheit. When we input a value, such as 0 (the freezing point of water in Celsius), the mapping provides us with the corresponding temperature in Fahrenheit, which is 32.

It is crucial that each input has only one output, although multiple inputs can yield the same output. This property ensures that a mapping is a precise and predictable process. Understanding this concept is fundamental in grasping more intricate aspects of function composition, where one mapping is applied after another.
Identity Mapping
An identity mapping, represented as 1_A where A is the domain, is a special kind of mapping. It does exactly what it sounds like鈥攊t maps every element to itself. So, if you have a set of numbers, say the set A includes numbers 1, 2, 3, identity mapping will associate 1 with 1, 2 with 2, and so on.

The identity mapping is an essential concept in function composition because it acts as a 'do nothing' operation. When you combine, or compose, any function f with an identity mapping, the original function f remains unchanged. This is because applying the identity mapping does not alter any elements in the set.

Understanding identity mapping is beneficial for students who struggle with the abstractness of function composition, as it simplifies the notion of what happens when you apply a function to an already defined set. It is a foundational element when exploring the properties and behavior of more complex functions.
Domain and Codomain
The terms domain and codomain are crucial for understanding mappings in mathematics. The domain of a function is the set of all possible inputs for which the function is defined. In other words, it's where the function starts; every element that can be considered an input for the function is part of the domain.

The codomain, on the other hand, is the set that contains all potential outputs of a function. However, it's broader than the 'range', which only consists of outputs that the function actually generates. To visualize this, consider a function that takes any whole number and squares it. The domain is all whole numbers, but while the codomain could be considered all real numbers, the range would only be non-negative whole numbers, since squaring never produces a negative result.

It's important to differentiate between domain and codomain when dealing with function composition because the output of one function, which becomes the input of another, must be compatible with the domain of the subsequent function. This understanding helps students check whether a composed function is properly defined by ensuring the output of one mapping fits into the domain of the next.

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Most popular questions from this chapter

Determine the number of different mappings from \((a)\\{1,2\\}\) into \(\\{1,2,3\\},(b)\\{1,2, \ldots, r\\}\) into \(\\{1,2, \ldots, s\\}\)

Prove Theorem \(5.11 .\) Suppose \(\operatorname{dim} V=m\) and \(\operatorname{dim} U=n .\) Then \(\operatorname{dim}[\operatorname{Hom}(V, U)]=m n.\)

Define \(F: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2}\) and \(G: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2}\) by \(F(x, y, z)=(2 x, y+z)\) and \(G(x, y, z)=(x-z, y)\) Find formulas defining the maps: (b) \(3 F\) (a) \(F+G\) (c) \(2 F-5 G\)

Show that each linear operator \(T\) on \(\mathbf{R}^{2}\) is nonsingular and find a formula for \(T^{-1}\), where (a) \(T(x, y)=(x+2 y, 2 x+3 y),\) (b) \(T(x, y)=(2 x-3 y, 3 x-4 y)\)

Let \(V\) be the vector space of real \(n\) -square matrices, and let \(M\) be a fixed nonzero matrix in \(V\). Show that the first two of the following mappings \(T: V \rightarrow V\) are linear, but the third is not: (a) \(T(A)=M A,\) (b) \(T(A)=A M+M A\), (c) \(T(A)=M+A\)
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