Prove that there exists a unique injective map $\gamma:X/_R\to Y$ with image equals $f(A)$ with $f=\gamma \circ \phi.$ So recently I have established an equivalence relation $R$ on $X$, with $xRx'$ iff $f(x)=f(x')$.
Let $\phi:X\to X/_R$ be the map of sets sending $x\mapsto \bar{x}$. Prove that there exists a unique injective map $\gamma:X/_R\to Y$ with image equals $f(A)$ with $f=\gamma \circ \phi.$
I'm struggling to understand as to what this question is asking. I know the definitions for what equivalence classes are, injective maps, quotients of sets, and image but what does it mean "with image $f(A)$ with $f=\gamma \circ \phi.$" Can anyone explain this part in a different way?
Also, any hints as to how to show the existence/uniqueness of such a map?
 A: The question is : find a map $\gamma:X/_R\to Y$ such that


*

*$\gamma$ is injective

*$\mathrm{Im}(\gamma) = f(A) = \mathrm{Im}(f)$

*$f = \gamma \circ \phi$


The first thing to do is to define $\gamma$ : for any $\overline x\in X/_R$, let $\gamma(\overline x) := f(x)$. Notice that $\gamma$ is well defined, because any element of $\overline x$ has the same image by $f$.
Then: 


*

*Let $\overline x$ and $\overline y$ in $X/_R$, such that $\gamma (\overline x) = \gamma (\overline y)$. Then, for any $x\in \overline x$ and $y\in \overline y$, we have $f(x) = f(y)$. Thus, $x R y$, and $\overline x = \overline y$ follows.

*Let $y\in \mathrm{Im}(\gamma)$. There exists $\overline x\in X/_R$ such that $\gamma(\overline x) = y$, ie for all $x\in\overline x$, $f(x) = y$. As equivalence classes are not empty, you have a $x\in X$ such that $f(x) = y$, thus $y\in f(A)$.
Reciprocally, let $y\in f(A)$. The, there exists $x\in X$ such that $f(x) = y$, and $\gamma(\overline x) = f(x) = y$.

*This one holds by definition of $\gamma$.
Now, suppose $\gamma_1$ and $\gamma_2$ are two solutions. By condition 3., you know that for all $x\in X$, $\gamma_1(\overline x) = f(x) = \gamma_2(\overline x)$. Thus, $\gamma_1 = \gamma_2$, and the unicity follows.
