I'm asking only to verify if I got this right, since I am only starting out with abstract algebra and discrete structures. My math background is also very poor, so please bear with me.
Given two algebraic structures $A$ and $A'$, with binary operators $\circ$ and $\circ '$ and neutral elements $e$ and $e'$, my task was to show that a homomorphism $h: A \longrightarrow A'$ exists.
What I did was set: for any $x \in A$ $h(x)= e'$. What followed is this:
Let $a,b,e \in A$ and $a',b',e' \in A'$. Let $h(a)$ and $h(b)$, as per definition, map to $e'$.
We observe the neutral elements and their definitions/axioms: $$a' \circ' b' = e' $$ and $$a \circ b = e.$$
Having seen this, we also observe the following: $$h(a) \circ' h(b) = h(a \circ b) = h(e) = e'.$$
Hence a morphism exists.
Is this enough? I still have to prove that if an isomorphism $h : A\longrightarrow A'$ exists, then there exist left and right inverse elements. Can't I derive this from the my earlier proof, if it's correct, by saying that the existence of the neutral element is enough to prove this?
I deeply apologize if I frustrate any of you.