Theorem:
Let $R, T$ be rings and $\varphi : R \rightarrow T$ be a surjective homomorphism. Then $\varphi(J(R)) \subseteq J(T)$ .

Proof:
We shall use the characterization of the Jacobson radical as the set of all $a \in R$ such that for all $r \in R$ , $1-ra$ is left invertible.

Let $a \in J(R), t \in T$ . We claim that $1-t\varphi(a)$ is left invertible:

Since $\varphi$ is surjective, $t=\varphi(r)$ for some $r \in R$ . Since $a \in J(R)$ , we know $1-ra$ is left invertible, so there exists $u \in R$ such that $u(1-ra)=1$ . Then we have $$ \varphi(u) \left( \varphi(1)-\varphi(r)\varphi(a) \right) = \varphi(u)\varphi(1-ra) = \varphi(1)=1 $$ So $\varphi(a) \in J(T)$ as required.

Theorem:
Let $R, T$ be rings. Then $J(R \times T) \subseteq J(R) \times J(T)$ .

Proof:
Let $\pi_1 : R \times T \rightarrow R$ be a (surjective) projection. By the previous theorem, $\pi_1(J(R \times T)) \subseteq J(R)$ .

Similarly let $\pi_2 : R \times T \rightarrow T$ be a (surjective) projection. We see that $\pi_2(J(R \times T)) \subseteq J(T)$ .

Now take $(a,b) \in J(R \times T)$ . Note that $a = \pi_1(a,b) \in J(R)$ and $b = \pi_2(a,b) \in J(T)$ . Hence $(a,b) \in J(R) \times J(T)$ as required.