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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 $?

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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.)

enter image description here

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, $$\langle (b,b')\rangle = \operatorname{ker}(g\circ f) \neq \operatorname{ker}(g) \vee\operatorname{ker}(f) =\langle (a,a')\rangle.$$

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