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Here is the question I am trying to solve (it is also mentioned in the answer of this link):

Proof of the existence of a well-defined function $\bar{f}$.

Let $X$ and $Y$ two sets and let $f:X\to Y$. Define for $x_1,x_2\in X$ a relation in $X$ as $x_1\sim x_2$ if $f(x_1)=f(x_2)$.

  1. Prove that this defines an equivalence relation on $X$.

Now you can talk about the quotient $X/\sim \quad = \{[x]:x \in X\}$, the set of the classes of equivalence.

Define $\bar{f}(x):X/\sim \quad \to Y$ as $\bar{f}([x]) = f(x)$. Since the definition uses an element of the class this could be ill-defined.

  1. Prove that this is well defined.

Now define $\pi:X \to X/\sim$ as $\pi(x) = [x]$.

Then

  1. $f = \bar{f}\circ\pi$

  2. $\bar{f}$ is injective

  3. $\pi$ is surjective

For the proof of 1) and 5) I have no problem in them.

For the proof of 2)

Assume that $[x_{1}] = [x_{2}]$ then $[f(x_{1})] = [f(x_{2})]$ but then what I do not know how to complete. Could anyone help me in that, please?

For the proof of 3)

I do not know how to do it. Could anyone help me in that, please?

For the proof of 4)

I know that it should be the reverse of 2)

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When you try to prove 2), you put square brackets around $f(x_1), f(x_2)$. These are incorrect (in fact do not mean anything as there is no equivalence relation defined on $Y$ at this point). It should be:


For the proof of 2)

Assume that $[x_{1}] = [x_{2}]$ then $f(x_{1}) = f(x_{2})$


From this it is clear that $\bar{f}([x])= f(x)$ is well defined e.g. it does not matter which representative of the class $[x]$ you select.

For part 3): $$\bar{f}\pi(x)=\bar{f}([x])=f(x),$$ by definition of the functions $\bar{f}$ and $\pi$.

For part 4): You are right it is the reverse of 2):

If $\bar{f}([x_1])=\bar{f}([x_2])$ then $f(x_1)=f(x_2)$ so $[x_1]=[x_2]$, by the definition of the equivalence relation.

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  • $\begingroup$ in the proof of part 2) should not $f(x_{1})$ and $f(x_{2})$ be inside equivalence classes or no and why? $\endgroup$
    – user778657
    Sep 22 '20 at 9:28
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    $\begingroup$ No. The point is that whichever element $x_2$ you choose to represent $[x_1]$, you will always have $f(x_1)=f(x_2)$ by definition of the equivalence relation. Thus the output of $\bar f$ applied to $[x_1]$ is precisely the element $f(x_1)$, not an equivalence class. (Technically you could define an equivalence relation on $Y$ by $y_1\sim y_2$ if and only if $y_1=y_2$, but the equivalence classes would be in bijective correspondence with $Y$, so you may as well work with $Y$ itself). $\endgroup$
    – tkf
    Sep 22 '20 at 10:51

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