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?



Reversible Rings
Following on from Martin Brandenburg's remark, to answer the question in the negative it is enough to find a ring $R$ with no zero-divisors which has a non-reversible 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 two-sided 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 two-dimensional 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 non-reversible. 


Indeed, if $e\_{ij} \in S$ are the standard matrix units, then $e\_{12}e\_{11} = 0$ whilst $e\_{11}e\_{12} \neq 0$.


