Let $B \subseteq C$ be an inclusion of finite dimensional (associative) algebras over a field $k$. Assume that $C$ is a free $B$-module. Let $\bigoplus\_i U\_i$ be 
a decomposition of $B$ into indecomposable $B$-modules. Let $\bigoplus\_j V\_{ij}$ be a decomposition of the extension $C \otimes\_B U\_i$ into indecomposable $C$-modules.


**Question**: Pick $u \in U\_i$ nonzero. Is there $\varphi\_u \in Hom\_B(U\_i,V\_{ij})$ such that $\varphi\_u(u)$ is nonzero? 


*Remark*: By Frobenius reciprocity, $Hom\_B(U\_i,V\_{ij}) \simeq Hom\_C(C \otimes\_B U\_i,V\_{ij})$ which is nonzero.  

In the semisimple case, Schur's lemma implies the existence of an injective map $\varphi$, which solves the problem for all $u$ nonzero. In the non-semisimple case, we cannot expect the existence of such an injective map, we are just asking about the existence of a map $\varphi\_u$ as above with $u \not \in ker(\varphi\_u)$. 


Assuming a positive answer, some conditions could be useless, but it is exactly the case I need.



A weak Schur's lemma for non-semisimple finite dimensional algebras
No, not necessarily.


Consider the case $B=kH$, $C=kG$ of finite group algebras over a field $k$, where $H\leq G$. I'll write $\downarrow$ and $\uparrow$ for restriction and induction. This case has a couple of simplifying features. Firstly, induction is both left and right adjoint to restriction. Secondly, $kH$ is selfinjective, and so the projective module $U\_i$ is also injective. If the answer to the question were "yes", then there would be an injection from $U\_i$ to a direct sum of copies of $V\_{ij}\!\downarrow$. But since $U\_i$ is injective, this would split, and so, by Krull-Schmidt, $U\_i$ would be a direct summand of $V\_{ij}\!\downarrow$.


Suppose that $U\_i$ is the projective cover of the simple $kH$-module $S$ and $V\_{ij}$ the projective cover of the simple $kG$-module $T$.


Then $V\_{ij}$ is a direct summand of $U\_i\!\uparrow$ if and only if
$$\text{Hom}\_{kG}(U\_i\!\uparrow, T)\cong\text{Hom}\_{kH}(U\_i,T\!\downarrow)$$
is nonzero, which is the case if and only if $T\!\downarrow$ has a composition factor isomorphic to $S$.


$U\_i$ does *not* occur as a direct summand of $V\_{ij}\!\downarrow$ if and only if
$$\text{Hom}\_{kH}(V\_{ij}\!\downarrow,S)\cong\text{Hom}\_{kG}(V\_{ij},S\!\uparrow)$$
*is* zero, which is the case if and only if $S\!\uparrow$ does *not* have a composition factor isomorphic to $T$.


So we just need to find simple modules $S$ for $kH$ and $T$ for $kG$ for which $S$ is a composition factor of $T\!\downarrow$ but $T$ is not a composition factor of $S\!\uparrow$.


This happens quite commonly, but $S$ can't appear in the head or socle of $T\!\downarrow$, or else there would be a map in one direction between $S$ and $T\!\downarrow$, and hence between $S\!\uparrow$ and $T$, and so $T$ would appear in the socle or head of $S\!\uparrow$. So $T$ has to be reasonably large.


One example that I happen to be familiar with is where $k$ is algebraically closed of characteristic two, $G=A\_5$ and $H=A\_4$. Then $kG$ has a $4$-dimensional simple projective module $T$ and $kH$ has three one-dimensional simple modules: take $S$ to be one of the non-trivial ones. Then $S$ and $T$ have the required properties.


