# Proof of the existence of a well-defined function $\bar{f}$(2)

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)

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.

• in the proof of part 2) should not $f(x_{1})$ and $f(x_{2})$ be inside equivalence classes or no and why?
– user778657
Sep 22 '20 at 9:28
• 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).
– tkf
Sep 22 '20 at 10:51