Proving Equivalence and Function Equality in Real Analysis

AI Thread Summary
The discussion revolves around a problem in Real Analysis involving functions and equivalence relations. The user seeks assistance with proving that f(x) equals h(g(x)), having already solved the first three parts of the problem. They clarify that E represents the set of equivalence classes for the function, emphasizing the requirement for functions to produce the same output for equal inputs. The user is confused about how to proceed with the proof, particularly in establishing the equality of the functions. The conversation highlights the complexities of function equality within the context of equivalence relations in Real Analysis.
SomeRandomGuy
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Hey guys, wasn't sure what forum to post this in. So if this is the wrong forum, I apologize. Anyway, I have a problem in Real Analysis that I can't quite get. Here it is:

Let f:A->B and R is a relation on A such that xRy iff f(x) = f(y).
a.) Prove R is an equivalence relation
b.) Show g:A->E is surjective
c.) Show h:E->B is injective
d.) Prove f(x) = h(g(x)).

I solved parts a, b, and c. My problem is part d... I don't even know where to begin. It just doesn't make sense to me when I think about it. Thanks for any help.

EDIT: I just realized I didn't put what E is. E is the equivalence classes on any particular element. So, it's the set of all equivalence classes for this function.
 
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For two functions to be equal, they have to send the same element to the same image.
 
Galileo said:
For two functions to be equal, they have to send the same element to the same image.

So are we showing that if f(x) = h(g(x)), then g(x) = x? Here is exactly what I have written so far:

"Proof: In order to show that two functions are equal, we must show that for any x in the domain, we will get the same output y in the codomain. So, if f(x) = x, then h(g(x)) = x as well. By the definition if being injective, x = g(x)."

I'm lost from there.
 
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