Lifting maps of quotient modules Today I tried to check this, but couldn't see how to do it. I think it is probably a standard result, but a brief check of Atiyah-Macdonald didn't yield anything, and I don't know what to google for. A reference is also appreciate.
Consider a associative, unital, $K$-algebra $R$ and a nontrivial ideal $\bar{R}$. Now consider two $R$ modules $B$ and $B'$. Also consider the quotients $\bar{B}=B/\bar{R}B$ and $\bar{B'}=B'/\bar{R}B'$.
If we have a map $\varphi:\bar{B}\to\bar{B}'$, is there some condition that we can lift to a map of $B\to B'$?
If the quotient map on $B'$ splits, we have this lift, but this is not iff.
Thanks in advance!
 A: Here is one way to look at this problem.  I will write $I$ for $\bar{R}$.
Firstly, $B/IB = (R/I) \otimes_R B$, that is, $B/IB$ is the image of $B$ under the induction functor (change of rings) $R$-mod $\to R/I$-mod.  This functor will be written $\uparrow$.  Frobenius reciprocity tells us
$$ \hom_{R/I}( B\uparrow, B'\uparrow) \cong \hom_R (B, B'\uparrow\downarrow) $$
where $\downarrow$ is restriction $R/I$-mod $\to R$-mod.  We'll make this an identification, so we consider $\phi$ as an element of the right hand side.  The question is then: "is $\phi$ in the image of $\hom_R(B,B') \to \hom_R(B,B'\uparrow \downarrow)$ ?" (the map on homs arises from $B' \to B\uparrow \downarrow$, the morphism given by the universal property of induction).
This stuff fits into the long exact sequence obtained by applying $\hom_R(B, -)$ to $IB' \to B' \to B'\uparrow\downarrow$.  We get
$$0\to \hom_R(B,  IB') \to \hom_R(B,  B') \to \hom_R(B,  B'\uparrow\downarrow) \stackrel{\omega}{\to} \operatorname{Ext}^1 _R(B, IB') \to \cdots $$
$\phi$ being in the image is equivalent to its being in the kernel of the connecting homomorphism $\omega$, which is multiplication by the short exact sequence above.  So one answer to your question is:  $\phi$ lifts if and only if it is killed by multiplication by the short exact sequence.
Really though, this is just a restatement of the problem.  Clearly $B$ being projective is enough to guarantee the lift exists, as is any assumption that makes that ext group $\operatorname{Ext}^1 _R(B, IB')$ vanish.  But for arbitrary $\phi$ I don't know general conditions other than the one above.  It's sufficient for the induction functor to be full.  $R/I$ being flat over $R$ (so induction is exact) certainly isn't enough.
