Understanding H&J's Section 3 on Invertible & One-to-One Functions: Q&A

[Math Amateur]

I am reading "Introduction to Set Theory" (Third Edition, Revised and Expanded) by Karel Hrbacek and Thomas Jech (H&J) ... ...

I am currently focused on Chapter 2: Relations, Functions and Orderings; and, in particular on Section 3: Functions

I need some help with H&J's depiction of invertible functions and their relationship to one-to-one functions ...H&J's section on invertible and one-to-one functions (including examples) reads as follows:View attachment 7588
View attachment 7589In the above text from Karel Hrbacek and Thomas Jech (H&J) we read the following:" ... ... 3.8 Theorem. A function is invertible if and only if it is one-to-one ... ... "Surely it is better if we define things so a function is invertible if and only it is a bijection ... and this, I think ist the usual approach ...

Why do MHB members think H&J define things this way ... what are the benefits of defining invertible functions the way H&J do ...

Any comments ... would like to know what readers think ... should I persevere with H&J as a text ...Peter

 

[castor28]

Hi Peter,

You are right: it is more correct to say that a function is invertible if it is a bijection. More precisely, a function (whose domain is not empty) has a left inverse iff it is injective, and a right inverse iff it is surjective.

The approach of the author makes me think of the typical high school exercise: "Find the domain and range of the function <some complicated expression>".
This is an abuse of language: formally, the domain and co-domain are part of the definition of a function; you define a function as $f : A \to B$, followed by the details of the specification. The question should really be: "Find the largest subset $D\subset\mathbb{R}$ that can be used as a domain for $f$, and find the corresponding range $f(D)$".

I think that, in this case, the author takes a similar approach. If a function $f$ is injective, it has an inverse $g$, but the domain of $g$ is up to you to discover (it will be the range of $f$ instead of its co-domain).

 

[Math Amateur]

Hi Peter,

You are right: it is more correct to say that a function is invertible if it is a bijection. More precisely, a function (whose domain is not empty) has a left inverse iff it is injective, and a right inverse iff it is surjective.

The approach of the author makes me think of the typical high school exercise: "Find the domain and range of the function <some complicated expression>".
This is an abuse of language: formally, the domain and co-domain are part of the definition of a function; you define a function as $f : A \to B$, followed by the details of the specification. The question should really be: "Find the largest subset $D\subset\mathbb{R}$ that can be used as a domain for $f$, and find the corresponding range $f(D)$".

I think that, in this case, the author takes a similar approach. If a function $f$ is injective, it has an inverse $g$, but the domain of $g$ is up to you to discover (it will be the range of $f$ instead of its co-domain).

Hi castor28 ...

Thanks for your post ...

... most thoughtful and helpful ...

Peter

 

[castor28]

Hi Peter,

An additional remark: it could be interesting to look at how the author defines the domain of a function. I would be tempted to think that the book focuses more on the practical aspects of calculus, but that does not agree well with the title of the book.

 

[Math Amateur]

Hi Peter,

An additional remark: it could be interesting to look at how the author defines the domain of a function. I would be tempted to think that the book focuses more on the practical aspects of calculus, but that does not agree well with the title of the book.

Hi Castor ...

Given your last post, you may be interested in H&J's introduction to functions (including of course the definition of domain of a function) ... so I am providing that as follows:View attachment 7590
View attachment 7591Notice that in the above text from H&J they note that since a function is a relation then the concept of the domain of a relation applies ...

So to get the definition of domain ... we need to look at the introduction to relations ... so I am providing the introduction to relations which includes the definition of "domain" ... H&J introduce relations as follows:... see 2.3 Definition for the definition of "domain" ... ... View attachment 7592
View attachment 7593As you can see from 2.3 Definition above (near the end of the above text ...... the domain of a relation $$R$$, $$ \text{ dom } R = \{ \ x \ \mid \ \text{ there exists } y \text{ such that } xRy \ \} $$Peter

 

[castor28]

Hi Peter,

With this definition, the statement about invertible functions is indeed correct.

This definition of a function corresponds more to a "partial function", like those used in the theory of computability, as shown by the remark "$F(a)$ is not defined if $a\notin\mbox{dom }F$", and the specific definition 3.3(a).

Although the concept of partial function has its uses, this is not the usual definition of a function in set theory. In particular, the axiom schema of replacement requires a "total function" (defined everywhere on its domain).