Let $R$ be a commutative ring, $M$ be a left $R$-module and $a\in R$. Then the scalar multiplication $f(m)=am$ is an $R$-module endomorphism called homothety. Suppose that $M=M_1\bigoplus M_2$ and $g:M_1\to M_1,m_1\mapsto rm_1$ is any homothety of $M_1$. Can $g$ be lifted to become a homothety of $M$. In other words, is there any endomorphism $h:M\to M$ such that $\bar{g}=g\circ h$ or $\bar{g}=h\circ g$ is a homothety of $M$, so that in this case we have $\bar{g}(m)=tm,$ for some $t\in R$ and for all $m\in M$?
I have been trying this in the following way: Define a projection $e:M\to M_1$ such that $\bar{g}=g\circ e$ is an endomorphism of $M$. Then for any $m=(m_1,m_2)\in M, \bar{g}(m)=g\circ e(m_1,m_2)=g(m_1)=rm_1$. However, this is not right since $rm_1\neq rm$, so it is not the scalar endomorphism. Is there a counter example to this problem?
To get this off the unanswered queue: fix $r\in R$ and let $g:M_1\to M_1$ be the homothety given $m_1\mapsto rm_1$. Then the map $f:M\to M$ defined by $m\mapsto rm$ is also a homothety, and it restricts to $g$ on the submodule $M_1\leqslant M$, as desired.