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.