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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

Arturo Magidin
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    "I think we need $f(x)$ to exist for every $x\in A$". That's part of the definition of "function". When you say that $f\colon A\to B$ is a function, you are saying that $f$ is defined on all of $A$, and that for each and every $x\in A$ there exists a $y\in B$ such that $f(x)=y$. – Arturo Magidin Dec 16 '21 at 17:36
  • You need conditions on $f$. Consider the case in which $A={0,1}$ and $B={a,b}$. If $f\colon A\to B$ sends $0$ to $a$ and $1$ to $b$, but $\text{~}$ identifies all elements of $A$ together, and $\text{^}$ is just the relation $a\text{^}a$ and $b\text{^}b$, then $g$ is not well defined, since $[0]{\text{~}}=[1]{\text{~}}$, but $[f(0)]{\text{^}}\neq[f(1)]{\text{^}}$. – Arturo Magidin Dec 16 '21 at 17:38
  • You need to study what is going on here more. Come up with specific examples like Arturo does in the second comment. For more examples, consider $A = B = \Bbb Z$, with the equivalence relations $\equiv \bmod 2$ and $\equiv \bmod 3$, and consider various functions $f$ on $\Bbb Z$, such as $f(x) = ax$ for different integers $a$, or $f(x) = x^2$. Which of these define maps $g$? Do $g([2])$ and $g([4])$ give the same result? (Note that $[2] = [4]$ in $\Bbb Z/(\equiv\bmod 2)$.) Come up with other equivalence spaces and functions until you are comfortable with what this definition of $g$ is doing. – Paul Sinclair Dec 17 '21 at 16:15

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(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 $f$.