If an algebra $A$ belongs to a variety and $\phi$ is a homomorphism then is $\phi(A)$ in the variety?

Question

My attempt: Let $A$ be an algebra that belongs to variety $V$ and $\phi$ be a homomorphism $\phi: A \rightarrow B$. Then for any term $t(x_1,x_2,...,x_n)$ and any elements $a_1,a_2,...a_n \in A$, we have $\phi(t(a_1,a_2,...a_n )) = t(\phi(a_1), \phi(a_2),...\phi(a_n))$. Where do I go from here?

Answer 1

What is a variety $V$? It's a set of algebras defined by a set of equations. How do we show that $\phi(A)\in V$? We show it satisfies all the equations defining $V$.

So suppose $t(x_1,\dots,x_n) = t'(x_1,\dots,x_n)$ is one of these equations. To show that $\phi(A)$ satisfies this equation, we need to show that for any $b_1,\dots,b_n\in \phi(A)$, $t(b_1,\dots,b_n) = t'(b_1,\dots,b_n)$.

Hint: For each $b_i$, pick a preimage $a_i$ in $A$, and use the fact that $A$ satisfies the equation, together with the relationship between homomorphisms and terms that you wrote down in the question.