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Let $f$ and $g$ be injective functions such that $f$ is a mapping from $A$ to $B$ and $g$ is a mapping from $B$ to $A$. Define an equivalence relation R on $A$ by $s_1= g \circ f (s_2)$ or $s_2= g \circ f (s_1)$. Define a similar equivalence relation on $B$. For $s \in A$, we write $[s]$ for the class in $A/R$. Similar for the class in $B/\bar R$.

Define a map $H : A/R \to B/\bar R$ by $H([s]) = [f(s)]$.

How can I prove that $H$ is well-defined, injective and surjective?

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    you haven't defined a map so how do we know? – 2017-02-28
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    ooops i forgot. sorry. i added it. – 2017-02-28
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    Have you considered that two equivalence classes that share an element are the same set? – 2017-02-28
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    I didn't get what you exactly mean. But i've been tackling with this problem for awhile (~10hours) and because that i started to get confused on definitions etc. This my progress so far, for well defined part, say $s_1=g \circ f ) ^n (s_2)$ (w.l.o.g) then by adding $f$ to this equation, and using associvity, i get $f(s_1)=(f \circ g) ^n f(s_2)$ and by adding $g$ i get $g \circ f (s_1) = (g \circ f) ^(n+1) (s_2) $ so $[s_1] = [s_2]$. I think this proves if we put the values in the function but not sure if this is a good proof or not – 2017-02-28
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    If $c\in [a]$ and $c\in [b]$ then you have $[a]=[b]$. Keep in mind also well-definedness and injectivity are the same problem in opposite directions. – 2017-02-28
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    I think I've done a similar thing as I wrote on the upper post but not sure if it's correct or not – 2017-02-28

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