Onto set mapping is the surjective set mapping, and into injective?

[mcastillo356]

TL;DR

I'm almost sure, but need to check it with the forum. It's about the first chapter of "Introduction to topology and modern analysis" I am reading online.

The textbook is being fine. I asked the forum for some introduction to topology, and decided to start with Simmon`s. This naive question is due to ignorance of the words into and onto, which I don't distinguish in Spanish. A quick browsing sugests I'm right.

 

[fresh_42]

Say, we have $f\, : \,A\longrightarrow B.$

$f$ is onto if it is surjective, i.e., $\forall\,b\in B\,\exists\,a\in A\, : \,b=f(a).$
$f$ is into if it is injective, i.e., $f(a)=f(a')\Longrightarrow a=a'.$

However, the English language is not always very exact when it comes to "in", "to", or "into". I have seen examples where into just meant $f(A)\subseteq B$ without the requirement of being injective. So, onto equals surjective is always true, into equals injective depends on the author. At least to my experience.

In German, we would use "auf" = "onto" = surjective, "in"="into"=injective, and "nach"="to"= $f(A)\subseteq B.$

 

[mcastillo356]

Say, we have $f\, : \,A\longrightarrow B.$

$f$ is onto if it is surjective, i.e., $\forall\,b\in B\,\exists\,a\in A\, : \,b=f(a).$
$f$ is into if it is injective, i.e., $f(a)=f(a')\Longrightarrow a=a'.$

Yes.

However, the English language is not always very exact when it comes to "in", "to", or "into". I have seen examples where into just meant $f(A)\subseteq B$ without the requirement of being injective. So, onto equals surjective is always true, into equals injective depends on the author. At least to my experience.

Simmons is being precise. Into means inyective.

In German, we would use "auf" = "onto" = surjective, "in"="into"=injective, and "nach"="to"= $f(A)\subseteq B.$

That's accurate, indeed.
Thanks, @fresh_42 !

 

[Gavran]

The textbook is being fine. I asked the forum for some introduction to topology, and decided to start with Simmon`s. This naive question is due to ignorance of the words into and onto, which I don't distinguish in Spanish. A quick browsing sugests I'm right.

To be on the set means occupying the whole set in a particular way and the word “onto” follows the definition of a surjective function at all.
To be in the set means occupying the part of the set but it does not mean that every element of the part of the set is occupied only once and this is the main reason for confusion. The word “into” does not completely follow the definition of an injective function.

 

[PeroK]

Whether you use into and/or onto is upto you!

 

[mcastillo356]

I'm working on the Schroether-Bernstein theorem, this is, the next step in the textbook "Introduction to topology and modern analysis", by Simmons. We assume that $f:\,X\rightarrow{Y}$ is a one-to-one mapping of $X$ into $Y$, and that $g:\,Y\rightarrow{X}$ is a one-to-one mapping of $Y$ into $X$. Our task is to produce a mapping $F:\,X\rightarrow{Y}$ which is one-to-one onto.
Apologies for my reaction to the previous post.

 

[fresh_42]

You need to understand the words. As I see it, Simmons uses the following terms:

$f\, : \,X\longrightarrow Y$ mapping

$X$ is the domain of $f.$

$f(X)$ is the range of $f.$

$f$ maps $X$ into $Y$ means $f(X)\subseteq Y.$

$f$ maps $X$ onto $Y$ means $f(X)= Y,$ surjective.

$f$ maps $X$ one-to-one into $Y$ means $x\neq x' \Longrightarrow f(x)\neq f(x'),$ injective.

$f$ maps $X$ one-to-one onto $Y$ means $f$ is injective, and surjective, i.e., bijective.

So we have two injective embeddings $f\, : \,X\hookrightarrow Y$ and $g\, : \,Y\hookrightarrow X$ and want to define a bijection $F\, : \,X \longrightarrow Y,$ i.e. a mapping that is also surjective.

 

[PeroK]

For example, $X = Y = \mathbb R$ and $f = g$ are both the exponential function. Then how to construct $F$?

 

[martinbn]

For example, $X = Y = \mathbb R$ and $f = g$ are both the exponential function. Then how to construct $F$?

In this case you take the identity map.

 

[mcastillo356]

For example, $X = Y = \mathbb R$ and $f = g$ are both the exponential function. Then how to construct $F$?

No idea. I can only attempt at it: $F$ could be the identity function? @martinbn says so.
Another doubt: The Axiom of Choice is needed?

 

[fresh_42]

No idea. I can only attempt at it: $F$ could be the identity function? @martinbn says so.
Another doubt: The Axiom of Choice is needed?

No. What do you get from $g\circ f$?

 

[PeroK]

No idea. I can only attempt at it: $F$ could be the identity function? @martinbn says so.
Another doubt: The Axiom of Choice is needed?

I looked it up. The AC is not needed.

 

[mcastillo356]

No. What do you get from $g\circ f$?

$\mathbb{R}^+$?

 

[fresh_42]

$\mathbb{R}^+$?


View attachment 365430

No. We have $f\, : \,X\longrightarrow Y$ and $g\, : \,Y\longrightarrow X.$ No real numbers anywhere.

I had $F=f\circ g\circ f$ in mind, but I'm not sure whether this is necessary. Try to figure out what you need, when you assume the existence of an element $y\in Y$ that is not in the range of $f.$

 

[PeroK]

No. We have $f\, : \,X\longrightarrow Y$ and $g\, : \,Y\longrightarrow X.$ No real numbers anywhere.

I had $F=f\circ g\circ f$ in mind

Does that work?

 

[fresh_42]

Does that work?

I don't know, I haven't solved it. The idea was born from the fact that the problem statement introduced $F.$ I thought $f$ would already do since $fg$ should be a permutation.

 

[PeroK]

I don't know, I haven't solved it. The idea was born from the fact that the problem statement introduced $F.$ I thought $f$ would already do since $fg$ should be a permutation.

I suspect the proof is non trivial!

 

[martinbn]

Does that work?

It need not work. The image will be contained in the image of $f$, which need not be onto.

 

[fresh_42]

I suspect the proof is non trivial!

Indeed. I looked it up. There is really some work to do. I quote:

This seemingly obvious statement is surprisingly difficult to prove.

 

[PeroK]

Indeed. I looked it up. There is really some work to do. I quote:

That's why i checked about the AC. Think about how awkward it is to get a bijection between $\mathbb R$ and $\mathbb R^2$.

 

[Gavran]

I'm working on the Schroether-Bernstein theorem, this is, the next step in the textbook "Introduction to topology and modern analysis", by Simmons. We assume that $f:\,X\rightarrow{Y}$ is a one-to-one mapping of $X$ into $Y$, and that $g:\,Y\rightarrow{X}$ is a one-to-one mapping of $Y$ into $X$. Our task is to produce a mapping $F:\,X\rightarrow{Y}$ which is one-to-one onto.

You already have the proof of the theorem on page 29 of the book. By the way, the more common statement of the Schroeder-Bernstein theorem is this: if there are injections f : A → B and g : B → A, then there is a bijection h : A → B.

 

[mcastillo356]

You already have the proof of the theorem on page 29 of the book. By the way, the more common statement of the Schroeder-Bernstein theorem is this: if there are injections f : A → B and g : B → A, then there is a bijection h : A → B.

Working hard on it.

 

[fresh_42]

Working hard on it.

Here is another source, in case this is easier.
https://web.williams.edu/Mathematics/lg5/CanBer.pdf

 

[mcastillo356]

Lemma 2 is difficult to me

Why $A=A_{0}\sim{B_{1}}\sim{A_{2}}\sim{B_{3}}\sim{A_{4}}\sim{\cdots}$?

Attempt
$A=A_{0}$, $B_{1}=f(A_{0})$, and so on. Why is this a consequence of injectivity?


Attempt
A set mapping is injective if

$$\forall{a,\,b}\in{X}\quad{f{(a)}=f{(b)}\Rightarrow{a=b}}$$

$A=A_{0}$, $B_{1}=f(A_{0})$, and so on follows from the definition of injectivity.

 

[fresh_42]

If we have an injective map $f\, : \,A\rightarrow B$ then its restriction on the image $f'\, : \,A \rightarrow f(A)$ is bijective.

 

[Gavran]

By the way, the more common statement of the Schroeder-Bernstein theorem is this: if there are injections f : A → B and g : B → A, then there is a bijection h : A → B.

The first statement of the theorem was published in 1887 by Georg Cantor.
If M and N are two such sets that components M' and N' can be separated from them, of which it can be shown that |M|=|N'| and |M'|=|N|, then M and N are equivalent sets.

 

[mcastillo356]

Here is another source, in case this is easier.
https://web.williams.edu/Mathematics/lg5/CanBer.pdf

Lemma 5 and Lemma 6 seem easy, but I've had to think about set partitions and equivalence relations.

Attempt

For any equivalence relation on a set $X$, the set of its equivalence classes is a partition of $X$. Conversely, from any partition $P$ of $X$, we can define an equivalence relation on $X$ by setting $x\sim{y}$ precisely when $x$ and $y$ are in the same part in $P$.

A bijection is an equivalence relation.

Lemma 5

$$A=\bar{A}\cup{\tilde{A}}$$

$$B=\bar{B}\cup{\tilde{B}}$$

Equivalence relation:

$$\tilde{A}=A_o^*\sim{B_1^*}\sim{A_2^*}\sim{B_3^*}\sim{A_4^*}\sim{\cdots}$$

$$\tilde{B}=B_0^*\sim{A_1^*}\sim{B_2^*}\sim{A_3^*}\sim{B_4^*}\sim{\cdots}$$

Set partition:

$$A\simeq{\tilde{A}}=\displaystyle\bigcup_{n\geq{0}}A_n^*\sim\displaystyle\bigcup_{n\geq{0}}B_n^*=\tilde{B}\simeq{B}$$

$$\bar{A}=\displaystyle\bigcap_{n\geq{0}}A_n$$

$$\bar{B}=\displaystyle\bigcap_{n\geq{0}}B_n$$