Set Theory: Suppose A is non-empty, onto, then...

Chris686

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I'm likely not following the standard protocol with this question.

I don't want the answer. I would like to figure it out on my own, but I'm not sure where to start. This is the question proposed by the professor:

Suppose A is non-empty, and that f: A -> B is onto.
How would you explain that there is a function g: A -> B, assuming that were true?

Now, if I'm reading this, I think that f = g, which is why this is confusing.

But... Thinking about any function, there could be infinitely many equations that are equal to each other.

Just, for example, let's say f(x) = x + 17 and g(x) = x(1 + x) - x^2 + 17
These two equations are equivalent.

I feel like I'm rambling on, but if someone could give me a small hint, I'd appreciate it. I'm not sure if I'm sniffing in the right direction or not.
 
I'm likely not following the standard protocol with this question.

I don't want the answer. I would like to figure it out on my own, but I'm not sure where to start. This is the question proposed by the professor:
Suppose A is non-empty, and that f: A -> B is onto.
How would you explain that there is a function g: A -> B, assuming that were true?

Having taught a course like this many times, I would bet you that your lecturer meant for you to construct a function \(\displaystyle g:B\to A\).

\(\displaystyle (\forall b\in B)(\exists a_b\in A)[f(a_b)=b]\) How can you use that to define \(\displaystyle g~?\)
 
If f is itself a function, we are done- take g= f. So the only problem occurs when f is a relation but not a function. And that simply means that we have at least one member of A that f maps to different members of B. So we get g by "pruning" f. That is if p is a point in A such that f(p) has several values, define g(p) to be one of those values.
 
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