Proof.
Since connectivity and truncatedness of maps of spaces are defined fiberwise, we may henceforth assume that \(C=*\). In this case, (1) and (2) become:
\(f \colon A \rightarrow B\) is a map from an \(n\)-connected space to an \((n+1)\)-connected space, then the fibers of \(f\) are \(n\)-connected.
If \(f\colon A \rightarrow B\) is a map from an \(n\)-truncated space to an \((n+1)\)-truncated space, then the fibers of \(f\) are \(n\)-truncated.
For \(n=-2\), these statements are obvious, for higher \(n\) they can be easily verified from the long exact sequence of homotopy groups associated to \(f\). ◻