I'm trying to show that any congruence relation $\equiv$ induces a homomorphism over an arbitrary vocabulary with operators $o_i$ of arity $a_i$ and a structure $\alpha$. I defined a new structure with a new interpretation of $o_i$
$\beta = \{\{y \mid y \equiv x\} \mid x \in \alpha \}$
$o_i(X_1, ..., X_{a_i}) = \{ o_i(x_1, ..., x_{a_i}) \mid x_1 \in X_1, ..., x_{a_i} \in X_{a_i} \}$
I want to show that the following function is a homomorphism: $\varphi(x) = \{ y \mid y \equiv x \}$. This is equivalent to showing the following:
$o_i(\varphi(x_1), ..., \varphi(x_{a_i})) = \varphi(o_i(x_1, ..., x_{a_i}))$
It should be clear that:
$o_i(\varphi(x_1), ..., \varphi(x_{a_i})) = \{ o_i(a_1, ..., a_{a_i}) \mid a_i \in \varphi(x_i) \} = \{ o_i(a_1, ..., a_{a_i}) \mid a_i \equiv x_i \}$
$\varphi(o_i(x_1, ..., x_{a_i}) = \{ y \mid y \equiv o_i(x_1, ..., x_{a_i}) \}$
Now I need to show that the above two sets are equal. It is clear that $o_i(\varphi(x_1), ..., \varphi(x_{a_i})) \subseteq \varphi(o_i(x_1, ..., x_{a_i})$ by the definition of a congruence, but I cannot prove the other direction unless $o_i$ interpreted in the original structure is surjective. Can anybody help me with this?
Denote by $[x]$ for the equivalence containing $x$, in the quotient set of $\alpha$ by $\equiv$ (since a congruence is an equivalence relation, we can take set-theoretic quotient). Note that your map above $\varphi$ is nothing but the quotient map $\varphi: x \longmapsto [x]$.
Then the interpretation you have given of $o_i$ in the quotient, is called the quotient algebra of $\alpha$ by the congruence $\equiv$, which is by definition:
$o_i([x_1],...,[x_{a_i}]) = [o_i(x_1,...,x_{a_i})]$ - $(*)$
To show this is well defined, we use the assumption of$\equiv$ being a congruence:
Suppose $x_1,...,x_{a_i},y_1,...,y_{a_i} \in \alpha$ are such that $[x_j] = [y_j]$ for $j = 1,2,3,...,a_i$. We need to show:
$[o_i(x_1,...,x_{a_i})] = [o_i(y_1,...,y_{a_i})] $
holds true. Since $[x_j] = [y_j]$ if and only if $x_j \equiv y_j$, we apply the fact that $\equiv$ is a congruence to get $o_i(x_1,...,x_{a_i}) \equiv o_i(y_1,...,y_{a_i})$, and therefore the above equation holds.
So this answers your questions.
Remark:
The definition of a congruence is so as to force the (set-theoretic) quotient map to be a homomorphism with respect to the interpretation $(*)$. To see this, just note that if $\equiv$ was an arbitrary equivalence relation, then $(*)$ is equivalent the statement that it is a congruence.