Prove that surjective homomorphisms preserve satisfaction of equations

Question

I'm currently studying universal algebra and I'm having trouble to prove the following statement:

Let $\Sigma$ be a many sorted signature and A and B be $\Sigma$-algebras. Prove that if there is a surjective homomorphism $h: A \rightarrow B$ and $A \vDash t \approx t'$, then $B \vDash t \approx t'$.

Just to make things clear:

• $t$ and $t'$ are both terms from the term algebra $T(\Sigma, X)$, where $X$ is a set of variables
• $A \vDash t \approx t'$ means that the equality $t \approx t'$ is valid in the algebra $A$. This in turn means that for every assignment $\alpha: X \rightarrow A$, we have $\bar{\alpha}(t) = \bar{\alpha}(t')$, where $\bar{\alpha}: T(\Sigma, X) \rightarrow A$ is the unique homomorphic extension of $\alpha$. Essentially, $\bar{\alpha}$ is the function that calculates the value of its parameter (a term) with regards to the assignment of variables $\alpha$.

Answer 1

For anyone that might end up struggling with this, I think I found a solution:

Let $\beta: X \rightarrow B$ be an assignment function. We want to prove that $\bar{\beta}(t) = \bar{\beta}(t')$. Notice that, given a function $\bar{\beta}$, we can define a new function $\gamma: T(\Sigma, X) \rightarrow B$ such that $\gamma(t) = h(\alpha(t))$, where $\alpha(t)$ is the element $a$ from $A$ such that $h(a) = \bar{\beta}(t)$ (we can do this because $h$ is surjective). It is very obvious that $\gamma = \bar{\beta}$. Other than that, since $A \vDash t \approx t'$, we know that $\alpha(t) = \alpha(t')$. Lastly, since $h$ is a well-defined function, we know that $h(\alpha(t)) = h(\alpha(t'))$, which means that $\bar{\beta}(t) = \bar{\beta}(t')$.

I don't know if every step I took was valid, but it's the best I could come up with.

Answer 2

Consider that for each sort $s$, the map $h_s : A_s \to B_s$ is surjective.

Fix a sequence of sorts $s_1, \ldots, s_n$.

The induced map $h_{s_1, \ldots, s_n} : A_{s_1} \times \cdots \times A_{s_n} \to B_{s_1} \times \cdots \times B_{s_n}$ defined by $h_{s_1, \ldots, s_n}(x_1, \ldots, x_n) = (h_{s_1}(x_1), \ldots, h_{s_n}(x_n))$ is surjective.

Now consider the variable context $\Delta = v_1 : s_1, \ldots, v_n : s_n$, and consider a sort $s$ and terms $t, t’ : s$ in context $\Delta$. These terms induce maps $A_t, A_{t’} : A_{s_1} \times \cdots \times A_{s_n} \to A_s$ and $B_t, B_{t’} : B_{s_1} \times \cdots \times B_{s_n} \to B_s$.

The fact that $h$ is a homomorphism means that $h_s \circ A_t = B_t \circ h_{s_1, \ldots, s_n}$ and $h_s \circ A_{t’} = B_{t’} \circ h_{s_1, \ldots, s_n}$.

The fact that $A \models t \approx t’$ means exactly that $A_t = A_{t’}$. From this, it follows that $B_t \circ h_{s_1, \ldots, s_n} = h_s \circ A_t = h_s \circ A_{t’} = B_{t’} \circ h_{s_1, \ldots, s_n}$.

Because $h_{s_1, \ldots, s_n}$ is surjective (and thus an epimorphism) and $B_t \circ h_{s_1, \ldots, s_n} = B_{t’} \circ h_{s_1, \ldots, s_n}$, it follows that $B_t = B_{t’}$. That is, $B \models t \approx t’$.