Let $A$ be a ring and $I_0\supset I_1\supset I_2\supset\ldots$ a decreasing sequence of ideals. Let $R_i=A/I_i$ be the factor ring of $A$ by the ideal $I_i.$ Let $f_i:R_{i+1}\rightarrow R_i$ be the canonical homomorphism. Denote by $R$ the inverse limite of the sequence $(R_i, f_i)$. Suppose that $R$ is a reversible ring, that is, for $a, b\in R$ we have $ab=0\Rightarrow ba=0$. It is true that each ring $R_i$ is also reversible?

Following on from Martin Brandenburg's remark, to answer the question in the negative it is enough to find a ring $R$ with no zerodivisors which has a nonreversible ring ($S$, say) as a proper homomorphic image. In a reversible ring, the right annihilator of an element is also its left annihilator, and as such, a twosided ideal. So any simple ring with zero divisors is not reversible. For a concrete example, we can take $R$ to be the universal enveloping algebra of $\mathfrak{sl}_2(\mathbb{C})$ and $S$ the ring of $2 \times 2$ matrices over $\mathbb{C}$. Since $R$ has an irreducible twodimensional module (the natural representation of $\mathfrak{sl}_2$), there is a surjection $R \twoheadrightarrow S$. Then $R$ is a domain, hence reversible, and $S$ is a simple ring with zero divisors, hence nonreversible. Indeed, if $e_{ij} \in S$ are the standard matrix units, then $e_{12}e_{11} = 0$ whilst $e_{11}e_{12} \neq 0$. 

