Show that $Con(A) \times Con(B)$ embeds into $Con(A \times B)$ Show that $Con(A) \times Con(B)$ embeds into $Con(A \times B)$.
Intuitively this makes a lot of sense but I am not sure how to define the map. 
$\alpha : Con(A) \times Con(B) \to Con(A \times B)$ where $(\theta_A, \theta_B) \mapsto \theta_{A \times B}$
I am not sure if this suffices. It seems to be the most natural since congruences in $A \times B$ will be acting component-wise.  From here it is enough to show the $(\theta_A, \theta_B) \subseteq (\sigma_A,\sigma_B)$ iff $\theta_{A \times B} \subseteq \sigma_{A \times B}$
$(\theta_A, \theta_B) \subseteq (\sigma_A,\sigma_B)  \iff \theta_A \subseteq \sigma_A \& \theta_B \subseteq \sigma_B  \iff ( a \theta_Aa'\implies a\sigma_Aa' ) \&( b \theta_Bb'\implies b\sigma_Bb' ) \iff (a,b)\theta_{A \times B}(a',b') \implies (a,b)\sigma_{A \times B}(a'b')  \iff \theta_{A\times B} \subseteq \sigma_{A \times B}$
and that is homomorphic:
$(\theta_A,\theta_B) \cap (\sigma_A,\sigma_B) = (\theta_A \cap \sigma_A, \theta_B\cap\sigma_B) = \theta_{A \times B} \cap \sigma_{A\times B}$ (more to show here but this is a sketch of the idea)
and 
$(\theta_A,\theta_B) \vee (\sigma_A,\sigma_B) = (\theta_A \vee \sigma_A, \theta_B \vee \sigma_B) = \theta_{A \times B} \vee \sigma_{A\times B}$ where the join of two congruences is the congruence generated by them; namely, by transitive closure.
 A: Define $\Phi : \mathbf{ConA}_1 \times \mathbf{ConA}_2 \to \mathbf{Con} (\mathbf{A}_1 \times \mathbf{A}_2)$ by making $\Phi : \langle \alpha_1, \alpha_2 \rangle \mapsto \alpha$, where
$$\langle a_1, a_2\rangle \,\alpha\, \langle b_1,b_2 \rangle \;\Longleftrightarrow\; \langle a_1, b_1 \rangle \in \alpha_1 \quad\text{and}\quad \langle a_2,b_2 \rangle \in \alpha_2.$$
It is straightforward to prove this gives a congruence relation on $\mathbf{A}_1 \times \mathbf{A}_2$, and also that $\Phi$ is an order-embedding, thus, an injective and isotone map.
This proves half of the statements about $\Phi$ being an homomorphism of lattices:
$$\Phi(\alpha_1 \cap \beta_1, \alpha_2 \cap \beta_2) \subseteq \Phi(\alpha_1,\alpha_2) \cap \Phi(\beta_1, \beta_2)$$
and
$$\Phi(\alpha_1,\alpha_2) \vee \Phi(\beta_1,\beta_2) \subseteq \Phi(\alpha_1 \vee \beta_1, \alpha_2 \vee \beta_2).$$
Now, let $\alpha = \Phi(\alpha_1,\alpha_2)$ and $\beta = \Phi(\beta_1,\beta_2)$, and let $\langle a_1,a_2 \rangle, \langle b_1,b_2 \rangle \in A_1 \times A_2$ be such that 
$$\langle\langle a_1, a_2 \rangle, \langle b_1,b_2 \rangle\rangle \in \alpha \cap \beta.$$
It follows that
$$a_1 \alpha_1 b_1, \quad a_2 \alpha_2 b_2, \quad a_1 \beta_1 b_1, \quad a_2 \beta_2 b_2,$$
whence $\langle a_1,b_1 \rangle \in \alpha_1 \cap \beta_1$ and $\langle a_2,b_2 \rangle \in \alpha_2 \cap \beta_2$.
Thus
$$\langle\langle a_1, a_2 \rangle, \langle b_1,b_2 \rangle\rangle \in \Phi(\alpha_1 \cap \beta_1, \alpha_2 \cap \beta_2),$$
and therefore, 
$$\Phi(\alpha_1 \cap \beta_1, \alpha_2 \cap \beta_2) = \Phi(\alpha_1,\alpha_2) \cap \Phi(\beta_1, \beta_2)$$
Now suppose $\langle\langle a_1, a_2 \rangle, \langle b_1,b_2 \rangle\rangle \in \Phi(\alpha_1 \vee \beta_1, \alpha_2 \vee \beta_2)$, that is, $\langle a_1,b_1\rangle \in \alpha_1 \vee \beta_1$ and $\langle a_2,b_2\rangle \in \alpha_2 \vee \beta_2$.
There exist $c_1, \ldots, c_n \in A_1$ and $d_1, \ldots, d_n \in A_2$ such that (we may suppose the sequences have the same length)
$$a_1 = c_1 \,\alpha_1\, c_2 \,\beta_1\, c_3 \,\alpha_1\, \cdots \,\beta_1\, c_n = b_1$$
and 
$$a_2 = d_1 \,\alpha_2\, d_2 \,\beta_2\, d_3 \,\alpha_2\, \cdots \,\beta_2\, d_n = b_2.$$
Thus, $\langle \langle c_{2k-1},d_{2k-1}\rangle,\langle c_{2k},d_{2k}\rangle \rangle \in \Phi(\alpha_1,\alpha_2)$ and $\langle \langle c_{2k},d_{2k}\rangle,\langle c_{2k+1},d_{2k+1}\rangle \rangle \in \Phi(\beta_1,\beta_2)$ (we may suppose $n$ is odd).
Hence, $\langle\langle a_1, a_2 \rangle, \langle b_1,b_2 \rangle\rangle \in \Phi(\alpha_1,\alpha_2) \vee \Phi(\beta_1,\beta_2)$, and therefore, 
$$\Phi(\alpha_1,\alpha_2) \vee \Phi(\beta_1,\beta_2) = \Phi(\alpha_1 \vee \beta_1, \alpha_2 \vee \beta_2).$$
