Prove that if a ring $R$ has IBN (invariant basis number), then so does $R/I$ for every proper two-sided ideal I.
My attempt: $R$ has IBN, then $R^m\cong R^n$ implies $m=n$. I wish to show that $(R/I)^m\cong (R/I)^n$ implies $m=n$. Should there is a map $R/I\rightarrow R$? then the basis of $(R/I)^m$ gives a basis of $R^m$.
The proposition is incorrect.
Take any ring $R$ with the IBN and any nonzero ring $S$ which does not have the IBN. Then $R\times S$ has the IBN, but $\frac{R\times S}{R\times \{0\}}\cong S$ does not.
The correct version of the proposition is that if $R/I$ is a nonzero quotient with the IBN, then $R$ also has the IBN. That is asked and answered elsewhere on ths site.
OR you could also say "if $R$ does not have the IBN, then none of its nonzero quotients $R/I$ has IBN either.