Surjectivity of canonical homomorphism between free objects in varieties of algebras

Question

Suppose $V' \subseteq V$ are varieties of algebras, and $F_{V'}(X)$ and $F_V(X)$ are free objects over some set $X$ in $V'$ and $V$, respectively. There are functions $\eta_V: X \to F_V(X)$ and $\eta_{V'}: X \to F_{V'}(X)$, which satisfy the familiar universal property.

If $h: F_V(X) \to F_{V'}(X)$ is the unique homomorphism satisfying $h\eta_V = \eta_{V'}$, is $h$ a surjection?

Answer 1

Yes. We have $h(\eta_V(x))=\eta_{V'}(x)$, and the elements $\eta_{V'}(x)$ generate $F_{V'}(X)$.

Answer 2

Given that $V' \subseteq V$, and if $\mathbf{T}(X)$ is the algebra of all the terms in $X$, $$\theta = \{ (p,q) \in T(X)^2 : V' \vDash p \approx q \}$$ is a congruence on $\mathbf{F}_V(X)$, and there is a homomorphism $\varphi : \mathbf{F}_V(X) \to \mathbf{F}_{V'}(X)$ given by $t \mapsto t/\theta$.
This homomorphism $\varphi$ is onto, by construction of $\theta$ and satisfies the equality of your homomorphism $h$.
Since there is only one such homomorphism, $\varphi = h$, that is, $h$ in onto.