Is $f(\operatorname{rad}A)\subseteq\operatorname{rad}B$ for $f\colon A\to B$ not necessarily surjective? If I have two $K$-algebras $A$ and $B$ (associative, with identity) and an algebra homomorphism $f\colon A\to B$, is it true that $f(\operatorname{rad}A)\subseteq\operatorname{rad}B$, where $\operatorname{rad}$ denotes the Jacobsen radical, the intersection of all maximal right ideals?
I can think of two proofs in the case that $f$ is surjective, but both depend on this surjectivity in a crucial way. The first uses the formulation of $\operatorname{rad}A$ as the set of $a\in A$ such that $1-ab$ is invertible for all $b\in A$, and the second treats an algebra as a module over itself, and uses the fact that the radical of $A$ as a module agrees with the radical of $A$ as an algebra, and is the intersection of kernels of maps onto simple modules; here the surjectivity is needed to make $B$ into an $A$-module in such a way that the radical of $B$ as an $A$-module is contained in the radical of $B$ as a $B$-module.
If a counter example exists, $A$ will have to be infinite-dimensional, as in the finite dimensional case all elements of $\operatorname{rad}A$ are nilpotent, and (I think, although I don't remember a proof right now, so maybe I'm wrong) that the radical always contains every nilpotent element.
This is my first question on here, so let me know if I should have done anything differently!

No, it is false that the Jacobson radical is sent to the Jacobson radical.
Take $A=K[X]_{(X)}$ [localization at  $(X)$],  $B=K(X)$ and the inclusion $f:K[X]_{(X)}\hookrightarrow K(X)$
Then $f(Rad(A))=f(XK[X]_{(X)})=XK[X]_{(X)} \nsubseteq Rad(B)=Rad(K(X))=(0)$
