Suppose that \(A \xrightarrow{f} B \xrightarrow{g} C\) are composable maps of spaces, and let \(n \geq -2\).
If \(g\) is \((n+1)\)-connected and \(gf\) is \(n\)-connected, then \(f\) is \(n\)-connected.
If \(g\) is \((n+1)\)-truncated and \(gf\) is \(n\)-truncated, then \(f\) is \(n\)-truncated.
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 \(f\). ◻
Original source: arXiv:2401.02956v2
Original source · 2401.02956v2