Let $A,B$ be finite dimensional $K$-algebras. A $K$-linear functor $T:\mathrm{mod}\,B \rightarrow \mathrm{mod}\,A$ is called a representation embedding if $T$ is exact, $T$ maps indecomposable modules in $\mathrm{mod}\,B$ to indecomposable modules in $\mathrm{mod}\,A$, and $T$ preserves isomporhisms (i.e. $T(X) \cong T(Y) \Rightarrow X \cong Y$).
A $K$-algebra $A$ is said to be of wild representation type if (and only if) for any finite dimensional $K$-algebra $B$, there exists a representation embedding $T:\mathrm{mod}\,B \rightarrow \mathrm{mod}\,A$.
Given a $K$-algebra $B$, denote by $\underline{\mathrm{mod}\,B}$ the stable module category of $B$. That is, the the category whose objects are the same as $\mathrm{mod}\,B$ with the projectives removed, and whose morphisms are the $K$-vector space $\underline{\mathrm{Hom}_B(M,N)} = \mathrm{Hom}_B(M,N)/\mathcal{P}(M,N)$, where $\mathcal{P}(M,N)$ is the space of morphisms $M \rightarrow N$ that factor through a projective $B$- module.
My question is as follows. Suppose $A$ and $B$ are finite dimensional $K$-algebras, $B$ is of wild representation type, and there exists a $K$-linear functor $T:\underline{\mathrm{mod}\,B} \rightarrow \mathrm{mod}\,A$ that satisfies the properties of representation embedding. Then is it necessarily true that $A$ is also a wild algebra?
It seems clear to me that $A$ is wild if there exists a representation embedding $T:\mathrm{mod}\,B \rightarrow \mathrm{mod}\,A$, but is the same necessarily true if all we know is the above?