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Problem: Let $\sim$ be an equivalence relation over a set $X$ and let $X / \sim $ be the corresponding quotient set. There is a function \begin{align*} q: X \rightarrow X / \sim \ : x \mapsto [x] \end{align*} which maps each element $x \in X$ to its corresponding equivalence class in $X / \sim$. This mapping is called the quotient map corresponding to $\sim$.

1) Show that $q$ is surjective.

2) Suppose that $f : X \rightarrow Y$ is a function with the property that \begin{align*} x_1 \sim x_2 \Rightarrow f(x_1) = f(x_2). \end{align*} Prove that there exists an unique function $\bar{f} : X/ \sim \rightarrow Y$ with the property that \begin{align*} f= \bar{f} \circ q. \end{align*}

Attempt at proof: For part 1) I reasoned as follows: Let $[x] \in X/ \sim$ be arbitrary. Then we have to show that there exists an element $x \in X$ such that $q(x) = [x]$. Since no equivalence class in $X / \sim$ is empty, there always exists an $x \in [x]$ for each $x \in X$. This proves that $q$ is surjective.

2) For this part, I'm not sure how to proceed. We need to construct the function $\bar{f}$ I think. So I would let $[x_1] \in X / \sim$. Then $\bar{f} [x_1] = y_1$ for some $y_1 \in Y$. Then we need to show somehow that $f = \bar{f} \circ q$ holds? How can I do that?

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    $\begingroup$ Part I is spot-on. $\endgroup$ Aug 7 '15 at 13:43
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For $[x]\in X/\sim$, define ${\overline f}([x]) = f(x)$. This gives $\overline{f}\circ q = f$. But it is dangerous, because it might not be well-defined.

However, suppose that $x_1\in[x]$. Then $x\sim x_1$. But your hypothesis implies that $f(x) = f(x_1)$. Peace now reigns in the valley. The proposed function, $\overline f$ is indeed a well-defined function.

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  • $\begingroup$ I see. It might not be well defined because the same $\bar{f}([x])$ might map to different elements? Ok, but then I don't understand the link between this and the second part of your argument. Why do you let $x_1 \in [x]$? $\endgroup$
    – Kamil
    Aug 7 '15 at 13:53
  • $\begingroup$ This shows that all elements of $[x]$ are mapped to the same place, so the value of $f(x)$ does not depend upon the choice of the element in $[x]$. $\endgroup$ Aug 7 '15 at 13:58
  • $\begingroup$ I see. And how do we prove the uniqueness of $\bar{f}$? $\endgroup$
    – Kamil
    Aug 7 '15 at 14:34
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Hint: Let's say that $f(x_1) = y_1$. This means that $\bar{f}(q(x_1))=y_1$. Can you use this to show what the function $\bar{f}$ does to an element of $X/\sim$?

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  • $\begingroup$ I would say that if $[x]$ is an element of $X / \sim$, then $\bar{f}$ maps every equivalence class to the elements in the image of $\bar{f}$ that are elements of the corresponding equivalence class? So I should define $\bar{f}([x]) = f(x)$? $\endgroup$
    – Kamil
    Aug 7 '15 at 13:59
  • $\begingroup$ @Kamil That's correct. Note that because $q$ is surjective, this completely defines $\bar{f}$ since we know the unique value of $\bar{f}([x])$ for every possible $[x]$. $\endgroup$
    – eigenchris
    Aug 7 '15 at 14:04
  • $\begingroup$ Does this prove the uniqueness of $\bar{f}$? $\endgroup$
    – Kamil
    Aug 7 '15 at 14:28
  • $\begingroup$ @Kamil Yes. If $f(x_1) = y_1$, then $\bar{f}$ has no choice in where it sends $[x_1]$; it is required that $\bar{f}([x_1]) = y_1$. We could try making up another function $\bar{g}$ with the property that $f = \bar{g} \circ q$ but we would again end up with $\bar{g}([x_1]) = y_1$, meaning $\bar{g} = \bar{f}$. $\endgroup$
    – eigenchris
    Aug 7 '15 at 14:42

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