Let $A, B$ be unitary rings and let $f$ be a surjective homomorphism of rings. Then $f(Jac(A)) \subseteq Jac(B)$ Let $A, B$ be unitary rings, and let $J(A)$, $J(B)$ denote the Jacobson's radical of $A$ and $B$. Let $f$ be a surjective homomorphism of rings. Then:
$$f(J(A)) \subseteq J(B)$$
Attempt: It suffices to prove that the preimage of every maximal ideal of $B$ is a maximal ideal in $A$. Let $M \subseteq B$ be an ideal of $B$, and let $f^{⁻1}(M)$ denote its preimage. Suppose $J$ is an ideal of $A$ such that $f^{⁻1}(M) \subseteq J$. Since $f$ is a surjection, this implies $M \subseteq f(J)$. Since $M$ is maximal, we have $f(J)=M$ or $f(J)=B$. If $f(J)=M$, then $J \subseteq f^{⁻1}(M)$ and then $J=f^{-1}(M)$. If $f(J)=B$, then I intended to prove that $1_{A}$ is in $J$, since any surjective homomorphism maps $1_{A}$ to $1_{B}$.The problem is that might be that exist other elements of $A$ mapped to $1_{B}$, and then we can't conclude that $1_{A} \in J$. 
My questions:
1.Is the initial problem true?
2.Is the preimage of a maximal ideal, through an epimorphism, also a maximal ideal?
3.Are there any ring homomorphisms $f:A\rightarrow B$ that maps $1_A$ and another element to $1_B$?

*

*Yes, it is true, for the reason below:

*Yes, this is true as well. By the correspondence theorem, the ideals of $ B $ correspond bijectively to the ideals of $ A $ containing $ \ker f $ in an inclusion respecting manner, so a maximal ideal of $ B $ pulls back to a maximal ideal of $ A $. Note that this is not true if we don't require the homomorphism to be surjective: if $ R $ is any commutative ring with a non-maximal prime ideal $ \mathfrak p $, then the preimage of the maximal ideal $ \mathfrak p R_{\mathfrak p} $ under the natural map $ R \to R_{\mathfrak p} $ is $ \mathfrak p $, which is not maximal.

*Sure, for example, the quotient map $ \mathbf Z \to \mathbf Z/2\mathbf Z $ is one such homomorphism.

