I am working on the following:
Problem: Let $(H, \cdot)$ be a semigroup and $\sim$ an equivalence relation on $H$. We say that $\sim$ is compatible with $(H, \cdot)$ $$: \iff \forall a,a',b,b' \in H: (a \sim a' \wedge b \sim b') \implies (a \cdot b \sim a' \cdot b') $$ Assume that $\sim$ is compatible with $(H, \cdot)$. Introduce $\bar{H}:= \lbrace [a] \mid a \in H \rbrace$ the set of all equivalence classes of $\sim$
Show that for $a,b \in H$ $$\bar{\cdot} : \begin{cases} \bar{H} \times \bar{H} &\longrightarrow \bar{H} \\ ( [a], [b]) & \longmapsto [a\cdot b] \end{cases} $$ defines a mapping
My approach: I am having a rough time with this one, but I believe I know what I have to do.
First I need to show that for arbitrary $[a],[b] \in \bar{H}$ I can always find $[a \cdot b]$ that is in $\bar{H}$, but I do believe that this is easily done due to the fact that $(H, \cdot)$ is a semigroup, thus closed under multiplication, therefore $ab \in H$ and $[ab] \in \bar{H}$ is that correct?
Secondly I need to show that if for some $([a], [b]) \in \bar{H} \times \bar{H}$ happens to have 2 pictures in $\bar{H}$ that they are in fact the same. Meaning $$ \bar{\cdot}([a],[b])=[a\cdot b]=[c \cdot d] \implies a\cdot b = c \cdot d$$ I am sure that I have to apply the definition of the $\sim$ equivalence relationship here that is compatible with $(H, \cdot)$ so I have $$[a\cdot b ] = [c \cdot d] \implies ab \sim cd $$ but I don't know how to continue from here, I haven't made use so far that $\sim$ is an equivalence relation, thus reflexive, symmetric and transitive, but I also don't see how to apply those properties.
Ps: I am afraid that I can't find good tags for this question, it is in fact out of an linear algebra class.
I'm going to use a different notation. $$ \newcommand{hbar}{\overline{H}} \newcommand{mbar}{\overline{m}} \newcommand{tim}{\widetilde{m}} \newcommand{hm}{\widehat{m}} $$
I'm going to use $p : H \to \hbar: x \mapsto [x]$ to denote the function that sends an element of $H$ to its equivalence class, and I'm going to use $m: H \times H \to H: (a, b) \mapsto a \cdot b$ to denote multiplication. I'll use $\mbar$ to denote the function we're trying to define (from $\hbar \times \hbar \to \hbar$).
Let me define a map from $H \times H \to \hbar$, by $$ \tim(a, b) = p(m(a, b)) = p(a \cdot b), $$ in short: $\tim$ takes the equivalence class of $a \cdot b$.
Let me show that $\tim$ can be turned into a function from $\hbar \times H \to \hbar$, via the rule $$ \hm :\hbar \times H \to \hbar : ([a], b) \mapsto [a \cdot b]. $$ This "definition" has to be shown to be independent of the choice of representative of $[a]$; in particular, if $a'$ is a different element of $[a]$, we need to show that $[a' \cdot b]$ is the same equivalence class as $[a \cdot b]$. To be more precise, we need to show that $p(m(a', b)) = p(m(a, b))$.
Your main hypothesis says that for any $s, t, s', t' \in H$, you have $$ (p(s) = p(s')) \wedge (p(t)= p(t')) \Longrightarrow p(m(s,t) = p(m(s', t')). $$
In the current context: let's apply that to the case $s = a, s' = a', t= b, t' = b$. Since we have $a \sim a' $, we have $p(s) = p(s')$. Since $b \sim b$ (reflexivity!), we have $p(t) = p(t')$. So we can conclude that $p(m(s, t)) = p(m(s', t'))$. So $\hm$ is now well-defined.
You now need to apply exactly the same reasoning to the right argument ($b$) to show that $\hm$ "passes to the quotient", i.e., can be used to define a function on $\hbar \times \hbar$ (namely $\tim$!) that has the required property. I'm going to let you give that a tr, and I'll critique whatever you do.
You'll notice that nothing at all in this argument depended on $m$ being multiplication in a semigroup. In fact, the general notion of "when does a function $f$ from $A$ to $B$ "pass to the quotient" to become a function from $A/\sim \to B$" is what you're really using here. And the answer is "whenever $f$ is constant on equivalence classes."