= Solution
If $X$ is <connected>, its <continuous> quotient image is <connected> by part (a). Conversely, suppose $X$ is disconnected. It has a nonempty proper subset $A$ both open and closed. Part (i) makes $A$ and its complement <saturation of a subset>[saturated], so $q^{-1}(q(A))=A$ and $q^{-1}(q(X\setminus A))=X\setminus A$. The <quotient topology> therefore makes these disjoint nonempty images open, and they cover $X/R$. Thus the quotient is disconnected. This proves \b[the two spaces are <connected> simultaneously].
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