Proving Left Inverse of Injective Function

[Bashyboy]

Homework Statement

$f : A \rightarrow B$ if and only if $\exists g : B \rightarrow A$ with the property $(g \circ f)(a) = a$, for all $a \in A$ (In other words, $g$ is the left inverse of $f$)

Homework Equations

The Attempt at a Solution

I have already prove the one direction. Now I am trying to verify that, if $f$ is injective, then there must exist a left inverse $g$. However, I am having great difficulty in doing so. I am understand the philosophy behind it: I have to define a function $g$ so that $(g \circ f)(a) = a$ holds true. But I cannot see how to explicitly use f's injectivity to construct such a function. Would someone mind pointing me in the right direction?

 

[pasmith]

Homework Statement

$f : A \rightarrow B$ if and only if $\exists g : B \rightarrow A$ with the property $(g \circ f)(a) = a$, for all $a \in A$ (In other words, $g$ is the left inverse of $f$)

Do you mean "f: A \to B is injective if and only if ..."?

Homework Equations

The Attempt at a Solution

I have already prove the one direction. Now I am trying to verify that, if $f$ is injective, then there must exist a left inverse $g$. However, I am having great difficulty in doing so. I am understand the philosophy behind it: I have to define a function $g$ so that $(g \circ f)(a) = a$ holds true. But I cannot see how to explicitly use f's injectivity to construct such a function. Would someone mind pointing me in the right direction?

Let b \in B. Either there exists an a \in A such that f(a) = b or there does not.

Suppose first that such an a exists. Since f is injective ...

 

[Bashyboy]

Because $f$ is injective, then if there exists another, call it $a' \in A$, so that $f(a') = b$, then we can conclude $a = a'$. Thus, $a$ is uniquely mapped to $b$.

 

[HallsofIvy]

Actually, it appears he is trying to prove "f is bijective".

 

[pasmith]

Actually, it appears he is trying to prove "f is bijective".

Mere injectiveity is necessary and sufficient for a left inverse to exist, which is what the OP is trying to show.

(Also, the title of the thread is "injective proof", not "bijective proof". Further, the OP didn't correct my assumption.)

 

[Bashyboy]

I apologize for not responding sooner. My idea was, that if $f(a)$ is uniquely associated to $a$, so that, if $g$ is defined to take back $f(a)$ back to $a$, then it is a well-defined function. Isn't that right?

 

[pasmith]

I apologize for not responding sooner. My idea was, that if $f(a)$ is uniquely associated to $a$, so that, if $g$ is defined to take back $f(a)$ back to $a$, then it is a well-defined function. Isn't that right?

That deals with the case where b = f(a) for some a \in A. You have now to deal with those b \in B for which there is no such a.

 

[Bashyboy]

That deals with the case where b = f(a) for some a \in A. You have now to deal with those b \in B for which there is no such a.

Okay, I suspect that it does not matter, that if I consistently chose the same element in $A$, say $a_0 \in A$, then $g$ will be the function after which I am seeking. In other words, if I choose $g$ to map all of those elements $b \in B$ which are not associated with any elements in $A$ by $f$ to the element $a_0$. In short, $g(b) = a_0$, if there does not exist an $a \in A$ such that $f(a) = b$.

 

[Bashyboy]

So, does the reasoning in my last post appear correct?

 

[geoffrey159]

Yes it does, in post #6 you have designed a left inverse function of f from $B \rightarrow A$, that you improved in post #8 to a left inverse mapping from $B \rightarrow A$.