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

Question

Given $f: X \rightarrow Y$ there exists a well-defined function $$\bar{f}: X /\sim \rightarrow Y,$$ $$[x] \mapsto f(x).$$

Such that $\bar{f}$ is injective and $f = \bar{f} \circ \pi,$ where $$\pi : X \rightarrow X /\sim, $$$$x \mapsto [x].$$

My thoughts:

1- I know that proving that a function is well defined means proving that its definition does not depend on the representative but I do not know how to implement that. Could anyone help me in doing so please?

2-Also, I do not know how to prove the existence of such function. Any hints will be appreciated.

Answer 1

The problem you're trying to state should go like this:

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 define 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.

2. Prove that this is well defined.

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

Then

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

4. $\bar{f}$ is injective

5. $\pi$ is surjective.

I hope that stating the problem properly helps you solve it. :)