Show that $U(R) \to U(R/J(R)) $ is an epimorphism. $\textbf{The question is as follows:}$


Show that the canonical homomorphism $R \to R/J(R)$ maps the group of units $U(R)$ onto the group of units $U(R/J(R))$.


$\textbf{Some observations:}$
We need to show that the following arguments are equivalent for ring $R$ and ieals $I$, $J$ and Jacobson radical $J(R):$
$ \rm 1)\quad J \subset J(R)$ $~~~~~~~~~~~~~~~~~~~~~$ i.e.  $ \rm J$ lies in every max ideal of $ \rm R$.
$ \rm 2)\quad 1 + J \subset U\,$ $~~~~~~~~~~~~~~~~~~~$ i.e.  $ \rm 1 + j$ is a unit for every  $ \rm j \in J$.
$ \rm 3)\quad I < 1 \Rightarrow I + J < 1\,$$~~~~~~~$ i.e. proper ideals survive in $ \rm R/J$.
This implies that all the nonunits in $ \rm R$ survive as nonunits in $ \rm R/J,$
so the only units in $ \rm R/J$ are those that are images of units in $ \rm R$ and we will be done!
Can someone help me in showing this equivalence?
Thanks!
 A: Any ring homomorphism $\;\phi:R\to S\;$ maps (left, right double-sided) units to (l,r,ds) units, since
$$u\in U(R)\implies\exists\,r\in R\;\;s.t.\;\;ru=1\implies1=\phi1=\phi(ru)=\phi(r)\phi(u)\implies\phi(u)\in U(S)$$
You can generalize the above to left/double-sided units and etc. Observe that I undertake implicitly the important agreement that homomorphism between unital rings maps one to one... Now, some highlights:
(1)$\to$(2): Suppose that for some $\;j\in J\;$ we have that $\;1+j\notin U(R)\implies 1+i\in M\;$ , for some maximal (left, right, two sided...) ideal $\;M\le R\;$ . Get now your contradiction
(2)$\to$(3): suppose now that $\;I\lneqq R\;$ but $\;I+J=R/J\;$ , then
$$\forall\,r\in R\;\exists\,i_r\in I,\,j_r\in J\;\;s.t.\;\;r=i_r+j_r$$
But there exists a maximal ideal $\;M_I\;$ s.t. $\;I\le M_I\;$ , and since $\;J\le M_I\;$  we get
$$r=i_r+j_r\in M_I+M_I=M_I$$
Since the above is true also for $\;r=u\in U(R)\;$ , get now your contradiction.
Added on request Suppose now (3) is true for some ideal $\;J\le R\;$ , but (1) isn't true. Then, there exists some maximal ideal $\;M\le R\;$ s.t. $\;J\nsubseteq M\;$ . This means that $J+M=R\;$ since $\;M\lneqq J+M\;$ ...and this contradicts (3) !
