A Well-Defined Bijection on An Equivalence Class

Question

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DATA:

Let $f:X\rightarrow Y$ be a surjective function. Define a relation $\sim$ on $X$ by $$a\sim b~\iff~f(a)=f(b).$$ Let $S=X/{\sim}$, namely let $S$ be the set of equivalence classes of elements of $X$ under the equivalence relation $\sim$. Define a function $q:X\rightarrow S$ by $$\forall~a\in X~q(a)=[a].$$ Lastly, define $\overline{f}:S\rightarrow Y$ by $$\forall~a\in X~\overline{f}([a])=f(a).$$

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QUEST:

$\dagger_1\hspace{0.5cm}$Is $\overline{f}$ well-defined?

$\dagger_2\hspace{0.5cm}$Is $\overline{f}$ a surjection? $\leftarrow$ Unsure about this one.

$\dagger_3\hspace{0.5cm}$Is $\overline{f}$ an injection?

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KNOWN:

$\overline{f}\circ q = f$ $\leftarrow$ What is the utility of this for the quest?

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DIAGRAM:

$\hspace{4.6cm}$

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THOUGHTS:

$\dagger_1^{\star}\hspace{.5cm}$"$\overline{f}$ well-defined": $ [a]=[b]\implies \overline{f}([a])=\overline{f}([b])$

$\dagger_2\hspace{.5cm}$"$\overline{f}$ surjection": Want to show that $\forall ~y\in Y~\exists~[x]\in S$ s.t. $\overline{f}([x])=y$ $\leftarrow$ ???

$\dagger_3^{\star}\hspace{.5cm}$"$\overline{f}$ injection": Want to show that $\overline{f}([a])=\overline{f}([b])\implies [a]=[b]$

$\star$ denotes the ones I think I've got so far.

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ATTEMPT:

$\dagger_1^{\star}\hspace{.5cm}$ $[a]=[b]\implies a\sim b \iff f(a)=f(b) \implies \overline{f}([a])=\overline{f}([b])$

$\dagger_2\hspace{.5cm}$ ...PENDING... $\leftarrow$ Unsure about this one.

$\dagger_3^{\star}\hspace{.5cm}$ $f([a])=f([b])\implies f(a)=f(b) \iff a\sim b \implies [a]=[b]$

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NOTES:

I suspect $q$ is to be used for the surjection proof.

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Answer 1

1. The answer is yes. Suppose $a\sim b$. Then $f(a) = f(b)$. So the map $[a]\mapsto f(a)$ is well-defined.

2. Suppose $f(a) = f(b)$. Then $a\sim b$ so $[a] = [b]$; $\overline f$ is 1-1.

3. It is easy to see that the image of X under $f$ is equal to that of the image of $X/{\sim}$ under $\overline f$. Therefore it is onto.

Answer 2

It is well defined because if you can any two representatives of a class, they have the same image by $f$.

It is a sujection because for any $y\in Y$, you have an $x\in X$ so that $f(x)=y$ because $f$ is surjective and that $x$ is in its own class.

It is an injection because if you have two elements $x_1$ and $x_2$ in $X$ whose class have the same image by $\overline{f}$, then $x_1$ and $x_2$ have the same image by $f$ so they are in the same class. Which means that is two classes have the same image by $\overline{f}$, they are equal.