Injective Proof

1. May 21, 2015

Bashyboy

1. The problem statement, all variables and given/known data
$f : A \rightarrow B$ 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 $f$)

2. Relevant equations

3. 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?

2. May 21, 2015

pasmith

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

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 ...

3. May 21, 2015

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 $b$.

4. May 22, 2015

HallsofIvy

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

5. May 22, 2015

pasmith

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.)

Last edited: May 22, 2015
6. May 22, 2015

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?

7. May 22, 2015

pasmith

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 $a$.

8. May 31, 2015

Bashyboy

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 $f(a) = b$.

9. Jun 6, 2015

Bashyboy

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

10. Jun 6, 2015

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 $B \rightarrow A$.