homomorphisms and congruence relations Do compositions of homomorphisms in universal algebra correspond to joins of congruence relations? That is- is the congruence relation $g \circ f(a ) = g \circ f( b) \Leftrightarrow a \sim b $ the join of $f(a ) = f( b) \Leftrightarrow a \sim b $ and $g (c) = g(d) \Leftrightarrow c \sim d $?
 A: The short answer: No.  
The long answer:
The congruences you are asking about are the kernels of $f$ and $g$ and $g\circ f$.  I'll denote these using $\operatorname{ker}$.  So, for example,
$$\operatorname{ker}(f) = \{(a,a')\in A^2 \mid f(a)=f(a')\}.$$
In general,
if $f \in \operatorname{Hom}(\mathbf{A}, \mathbf{B})$ and 
$g \in \operatorname{Hom}(\mathbf{B}, \mathbf{C})$,
 then the kernel of $g$ is a subset of $B^2$ while the kernel
 of $f$ is a subset of $A^2$, so these are not even in the same congruence lattice, and the join 
$\operatorname{ker}(g) \vee \operatorname{ker}(f)$ 
doesn't make sense.  However,
we can make sense of your question if we assume
that
$f\in \operatorname{End}(\mathbf{A})$ 
and
$g \in \operatorname{Hom}(\mathbf{A}, \mathbf{B})$.  Then
$\operatorname{ker}(g)$ and $\operatorname{ker}(f)$ both belong to 
the congruence lattice of $\mathbf{A}$ (denoted
$\operatorname{Con}(\mathbf{A})$). 
Under these restricted conditions, your question makes sense and the answer is
no, it is not true in general that 
$$\operatorname{ker}(g\circ f) = \operatorname{ker}(g) \vee\operatorname{ker}(f).$$
Here is a counterexample: Let the algebra $\mathbf{A}$ be the lattice shown on
the left in the diagram below.  (The congruence lattice of $\mathbf{A}$ is shown on the right.)  

Now define
$f\in \operatorname{End}(\mathbf{A})$ by 
$f(0) =0$,
$f(1) = 1$,
$f(a) = b$,
$f(a') = b'$,
$f(b) = a$,
$f(b') = a'$
(so, in fact, $f\in \operatorname{Aut}(\mathbf{A})$, but no matter).
Define
$g \in \operatorname{End}(\mathbf{A})$ as follows: 
$g$ acts as the identity on all elements of $\mathbf{A}$ except $a'$, where 
$g(a') = a$.  Then it's
easy to check that
$ \operatorname{ker}(f) = 0_\mathbf{A}$,
$\operatorname{ker}(g) = \langle (a,a')\rangle$, and
$\operatorname{ker}(g\circ f) = \langle (b,b')\rangle$.
Therefore,
$$\langle (b,b')\rangle = \operatorname{ker}(g\circ f) \neq \operatorname{ker}(g)
\vee\operatorname{ker}(f) =\langle (a,a')\rangle.$$
