Chapter 3: Problem 2
Show that if \(f: A \rightarrow B\) has an inverse \(g,\) then \(f\) is bijective.
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none}
Learning Materials
Features
Discover
Chapter 3: Problem 2
Show that if \(f: A \rightarrow B\) has an inverse \(g,\) then \(f\) is bijective.
These are the key concepts you need to understand to accurately answer the question.
All the tools & learning materials you need for study success - in one app.
Get started for free
Prove Proposition 3.16. That is, show that if \(f: A \rightarrow B\) is bijective, an inverse \(g\) of \(f\) exists. You have to define such a \(g\), show that it is a function, and show that it is an inverse of \(f\), i.e., \(f(g(y))=y\) and \(g(f(x))=x\) for all \(x \in A\) and \(y \in B\).
Suppose \(f: A \rightarrow B\) and \(g: B \rightarrow C\). Show that the graph of \(g \circ f\) is \(R_{f} \mid R_{g}\)
Given \(f: A \rightarrow B,\) define the partial function \(g: B \rightarrow A\) by: for any \(y \in B,\) if there is a unique \(x \in A\) such that \(f(x)=y,\) then \(g(y)=x ;\) otherwise \(g(y) \uparrow .\) Show that if \(f\) is injective, then \(g(f(x))=x\) for all \(x \in\) \(\operatorname{dom}(f),\) and \(f(g(y))=y\) for all \(y \in \operatorname{ran}(f)\)
Show that if \(f: A \rightarrow B\) and \(g: B \rightarrow C\) are both surjective, then \(g \circ f: A \rightarrow C\) is surjective.
Show that if \(f: A \rightarrow B\) and \(g: B \rightarrow C\) are both injective, then \(g \circ f: A \rightarrow C\) is injective.
What do you think about this solution?
We value your feedback to improve our textbook solutions.