Topology by Simmons Problem 1.3.3

[Figaro]

Homework Statement

Let ##X## and ## Y## be non-empty sets, ##i## be the identity mapping, and ##f## a mapping of ##X## into ##Y##. Show the following

a) ##f## is one-to-one ##~\Leftrightarrow~## there exists a mapping ##g## of ##Y## into ##X## such that ##gf=i_X##
b) ##f## is onto ##~\Leftrightarrow~## there exists a mapping ##h## of ##Y## into ##X## such that ##fh=i_Y##

Homework Equations

The Attempt at a Solution

a) ##~\Rightarrow~## Since ##f## is one-to-one, there exist ##y \in Y## such that ##y=f(x)## for some ##x\in X##, this shows that there exist at least some mapping ##g## that maps ##Y## into ##X## such that ##x=g(y)=g(f(x))=(gf)(x)## for some ##y##, and also ##gf=i_X##.

b) ##~\Rightarrow~## Since ##f## is onto, for all ##y\in Y## there exist some ##x\in X## such that ##y=f(x)##, this shows that there exist some mapping ##h## of ##Y## into ##X## such that ##x=h(y)## and ##y=f(x)=f(h(y))=(fh)(y)## which implies ##fh=i_Y##.

I have done the forward proof but I just want to know if my proof here is correct.

 

[Dick]

Homework Statement

Let ##X## and ## Y## be non-empty sets, ##i## be the identity mapping, and ##f## a mapping of ##X## into ##Y##. Show the following

a) ##f## is one-to-one ##~\Leftrightarrow~## there exists a mapping ##g## of ##Y## into ##X## such that ##gf=i_X##
b) ##f## is onto ##~\Leftrightarrow~## there exists a mapping ##h## of ##Y## into ##X## such that ##fh=i_Y##

Homework Equations

The Attempt at a Solution

a) ##~\Rightarrow~## Since ##f## is one-to-one, there exist ##y \in Y## such that ##y=f(x)## for some ##x\in X##, this shows that there exist at least some mapping ##g## that maps ##Y## into ##X## such that ##x=g(y)=g(f(x))=(gf)(x)## for some ##y##, and also ##gf=i_X##.

b) ##~\Rightarrow~## Since ##f## is onto, for all ##y\in Y## there exist some ##x\in X## such that ##y=f(x)##, this shows that there exist some mapping ##h## of ##Y## into ##X## such that ##x=h(y)## and ##y=f(x)=f(h(y))=(fh)(y)## which implies ##fh=i_Y##.

I have done the forward proof but I just want to know if my proof here is correct.

I'm sure you have the right idea. In the injective case you can simply define what ##g## is. For the surjective case you need to show there exists such a function ##h##. What you need for a formal statement is the axiom of choice.

 

[Figaro]

I'm sure you have the right idea. In the injective case you can simply define what ##g## is. For the surjective case you need to show there exists such a function ##h##. What you need for a formal statement is the axiom of choice.

So you mean in a) I should state it as,
Since ##f## is one-to-one and suppose that there exist a mapping ##g## of ##Y## into ##X##, then ##x=g(y)=g(f(x))=(gf)(x)## which implies ##gf=i_X##.

In b) didn't I already show that there exist a mapping ##h## since for every ##y## there is always an ##x## such that ##y=f(x)##? If not, what do you mean by using the axiom of choice? How should I use it to formalize the proof?

 

[Dick]

So you mean in a) I should state it as,
Since ##f## is one-to-one and suppose that there exist a mapping ##g## of ##Y## into ##X##, then ##x=g(y)=g(f(x))=(gf)(x)## which implies ##gf=i_X##.

In b) didn't I already show that there exist a mapping ##h## since for every ##y## there is always an ##x## such that ##y=f(x)##? If not, what do you mean by using the axiom of choice? How should I use it to formalize the proof?

Well, no. You need to define what the mapping ##g## is for ALL ##y## in ##Y##. You can't just suppose there is one. For the onto case you need to do the same thing for ##h##. If these are infinite sets you'll find you may have to make an infinite number of choices. The assumption you can do this is called the axiom of choice. It's a little technical and if you haven't talked about it just assume you can do that.