0
$\begingroup$

Question: Suppose that $g: X \to Y$ and $h: Y \to Z$ and $f: Y \to Z$ are functions so that $h \circ g = f \circ g$ and $g$ is surjective. Prove that $h = f$.

Comments:

Still trying to wrap this around my head… So basically the map should go: $$ X \to Y \to Z $$

$g$ is the step (function) of $X \to Y$.

$h$ and $f$ are the step (function) of $Y \to Z$.

In some ways I ignore $Y$ since it is an intermediate step and assume that elements of $Y$ such as $h \in Y$ and $f \in Y$ are equivalent to $X$. This should mean that $h$ and $f$ are logically equivalent to $g(X)$ where $g \in X$.

However, $h \circ g$ map would be $Y \to Z \to X \to Y$? But simplifying this should give $Z \to X$ because $Y$ comes from $X$ (I believe not sure). In the other case, $f \circ g$ map would be $Y \to Z \to X \to Y$? $g$ is surjective.

I know there’s something different between $h$ and $f$ but how does surjectivity make them not equal to each other and prove $h = f$? Both functions have the same $Y \to Z$ pattern where the $Y$ comes from the same $X$ (right?). It has something to do with surjectivity changing the property?

I know that surjectivity means: A function $f: A \to B$ is surjective iff $\forall b \in B \exists a \in A$ such that $f(a) = b$.

“Every $b \in B$ is a value of $f$.”

How does this make $h = f $illogical and not equal to each other?

I know that $h$ and $f$ can't be equal since each one has different domains and co-domains.

Any idea for explaining this and helping wrap up my thoughts on this problem would be highly appreciated. I've thought about this problem for a while and can't just seem to connect the dots between the map and $g$ being surjective to show $h = f$ (if possible).

  • 0
    The part quotes as "question" is completely different from the question title. Please fix them so they match. – 2017-02-10
  • 0
    Your $h$ should map $Y \to Z$ not $X \to Y$, otherwise the composition $h\circ g$ doesn't make sense. – 2017-02-10
  • 0
    Just fixed that @HenningMakholm – 2017-02-10
  • 0
    I thought it went from h ∘ g map would be Y → Z → X → Y since h is = Y → Z and g = X → Y. I thought this was the appropriate way to put it together the map. But now that I think about it h ∘ g means h(g(x)) which means Y → Z = (X → Y) → Z???? Not sure @m_t_ – 2017-02-10
  • 0
    Would it be OK if I edited your question to make the LaTeX right and to get the domains/codomains (X, Y, Z) so they make sense? – 2017-02-10
  • 0
    Sure I don't mind @JohnHughes – 2017-02-10
  • 0
    Done. See additional notes in my answer. – 2017-02-10
  • 0
    Thank you for your help. I think that this is one of the challenges of taking a logic or proof-based class. Totally different from calculus. It is hard to articulate things in such manner but you are right and this is a good mental exercise. @JohnHughes – 2017-02-10

1 Answers 1

1

Extended hint:

OK. To show $h = f$, you need to show that for every $y \in Y$, we have $$ h(y) = f(y). $$

That may be the biggest step in the whole proof: figuring out what you need to actually show, so look carefully and try to understand why that's how you prove two functions are equal.

To prove this, suppose that $y$ is any element of $Y$. Then because $g$ is surjective, there's an $x \in X$ with $g(x) = y$. What can you say about $$ h(g(x)) $$ and $$ f(g(x))? $$

Where does that lead you?

Additional notes: What you wrote in your question shows that you're not really understanding the notation $f: X \to Y$; that notation doesn't mean that the formula for $f$ is something like $f(x) = y$. It means that $f$ takes things in $X$ and produces things in $Y$. It's quite possible to have $u : X \to Y$ and $v: X \to Y$, without $u$ and $v$ being the same. For instance, with $X$ and $Y$ both being the set of real numbers, we could have $u(x) = x + 1$ and $v(x) = x - 1$. Since $u(2) \ne v(2)$, the function $u$ is different from the function $v$.

As for "I know there's something different between $h$ and $f$," that doesn't quite make sense; you are, after all, supposed to be proving that they're the same.

  • 0
    Show h = y. Let all y ∈ Y, then we get h(y) = f(y). Since there are y elements of Y and g is surjective, then there's an x ∈ X and this means g (x) = y. If this statement is true, then this should lead h(g(x)) = z? and f(g(x)) = x? Like are you saying that the variable will rotate from the map that was established? Not sure still but I guess that if it is equal both sides should end with the same answer. However, I don't think this is true and should give different variables y and z? Y is an intermediate step so is it skipped? Not sure if what I'm saying makes much sense. – 2017-02-10
  • 0
    I think that if I were to draw it X --> Y --> Z. I know that to get g(x) = y, I had to go from Y intermediate to X because of subjectivity. From here I think that maybe, h and f might be injective? that is the only thing that could keep it equivalent to each other? Again not sure what it leads me to but I suppose it should bring me back to Z step which has z elements of Z? – 2017-02-10