If I have two algebras $A,B$, and one is the quotient of the other, i.e. there exists a surjective morphism $\phi : A \to B$. Then is $B$ isomorphic to some subalgebra of $A$? I think so, because I just need to select for each equivalence class $\phi^{-1}(b)$ one element $a \in \phi^{-1}(b)$, and furthermore if I had selected $a_1 \in \phi^{-1}(b_1), a_2 \in \phi^{-1}(b_2)$, and $a_3 \in \phi^{-1}(b_1 \cdot b_2)$, then $a_3 = a_1 \cdot a_2$ must hold. Which could be realized I think.
On the other hand, if I have a subalgebra $A'$ of $A$, could it always be realisied as a quotient, i.e. is there some surjective $\psi : A \to A'$. I think this should be possible. But then if quotients and subalgebras are so related, then they are in some sense the same concepts, but as I see it they are threated quite differently in textbooks and theorems, so they are not the same, which make me doubt my constructions. Could someone please clarify?
The answer is NO of course. Take for example any algebra $A$ with has no 1-element subalgera and take their 1-element quotient $B$.