Let $A,B$ sets and $\text{~}, \text{^}$ equivalence relations in $A$ and $B$ respectively.
Let $f:A\rightarrow B$ a function. Find a condition that must be met for the function $$g:(A/_\text{~})\rightarrow (B/_\text{^})$$ defined by $g([x]_{\text{~}})=[f(x)]_{_\text{^}}$ to exist.
Find a condition for $g$ to be inyective.
My attempt: I think we need $f(x)$ to exist for every $x\in A$, so that $g([x]_{\text{~}})=[f(x)]_{\text{^}}$ could actually exist. If this does not happen, $f(x)$ could not exist for some $x\in A$ and that implies $[f(x)]_{_\text{^}}=\emptyset$. Is this good?
For the second part, I have to prove that $g$ is injective, which means that $g([a]_{\text{~}})=g([b]_{\text{~}})$ implies $[a]_{\text{~}}=[b]_{\text{~}}$. In this case, I think $f$ must be injective since $f(a)=f(b) \implies a=b$. Then $a\text{~}b$, which means that $[a]_\text{~}=[b]_\text{~}$. In consequence, $g$ is injective.
Can someone help me? Thanks
(Throughout this answer, I'll use "$\cong$" in place of your "$\text{^}$".) The definition of $g$ depends on the arbitrary choice of the representative $x$ in the class' representation $[x]_\sim$. So, in order to have a good definition, we need that $[y]_\sim = [x]_\sim\Longrightarrow g([y]_\sim)=g([x]_\sim)\stackrel{(g's\space def.)}{\iff}[f(y)]_\cong =[f(x)]_\cong$, namely: $$y\sim x\Longrightarrow f(y)\cong f(x) \tag 1$$ The condition for $g$ to be injective is precisely the other way around. In fact, $g([y]_\sim)=g([x]_\sim)\stackrel{(g's\space def.)}{\iff}$ $[f(y)]_\cong =[f(x)]_\cong\Longrightarrow [y]_\sim=[x]_\sim$, namely: $$f(y)\cong f(x) \Longrightarrow y\sim x \tag 2$$ Therefore, the existence of such a $g$, and its possible injectivity, depend on $f$.