Making a map injective by quoetienting by a certain subset. Take sets $X$ and $Y$ and a map $f:X\to Y$.
Let 
$$A=\{x\in X\;|\; \exists x'\not =x,\; f(x')=f(x)\}$$
Let $$\tilde f:X/A\to Y$$ defined by $$\tilde f([x])=f(x)$$ where $[x]$ denotes the equivalence class of $x$.
we claim that the map $\tilde f$ is injective.
Indeed, if $f([x])=f([y])$ then $f(x)=f(y)$ hence
$x$ and $y$ belong to $A$ so $[x]=[y]$ thus $\tilde f$ is injective.
and this gives a procedure how to make any map injective. Is my reasoning correct?
 A: Instead of writing $X/A$, we can define a relation $R$ on X such that $x=y$ iff $f(x)=f(y)$
This is an equivalence is very clear. 
Show we can talk of $X/R := $ the set of equivalence class of X. 
Define, $\tilde{f}:X/R\rightarrow Y$ as $\tilde{f}([x])=f(x)$. 
This map is clearly injective. And $X\rightarrow X/R$ there is natural map, you can think of $X/R$ as $X$ mod some subset of $X$.
Abstract Non-Sense: A map fails to be injective when two different element get mapped to a single element. If you want to make the map injective, you need to forcefully make those two different element a single element of the set. Going modulo equivalence does exactly this job.  
A: Your function $\bar f$ is not well-defined.
Let $f:\{0,1,2,3\}\rightarrow\{0,1\}$ prescribed by $0,1\mapsto0$ and $2,3\mapsto1$. 
From $f(0)=f(1)$ it follows that $0,1\in A$ and from $f(2)=f(3)$ it follows that $2,3\in A$.
Then $A=\{0,1,2,3\}$ so that:
$\bar f(A)=\bar f([0])=f(0)=0$ and $\bar f(A)=\bar f([2])=f(2)=1$
This shows that $\bar f$ is not well-defined.
