Homotopy group objects detect truncation and connectivity

(1) We first prove, by induction on \(r\geq -1\), that an \(r\)-truncated morphism has trivial \(\pi_k\) for \(k>r\). If \(r=-1\), then \(\Delta_f\) is an isomorphism, so \(\pi_k(f)\cong\pi_{k-1}(\Delta_f)=*\) for \(k\geq 1\). Moreover, since \(f\) is \(0\)-truncated, \(\pi_0(f)\) is represented by the projection \(X\times_YX\to X\), which is an isomorphism because \(f\) is a monomorphism. For the induction step, \(\Delta_f\) is \((r-1)\)-truncated, and hence for \(k>r\) by the induction hypothesis.We next prove the sharper assertion that if \(r\geq0\), the morphism \(f\) is \(r\)-truncated, and \(\pi_r(f)=*\), then \(f\) is \((r-1)\)-truncated. For \(r=0\), the preceding description identifies \(\pi_0(f)\) with \(X\times_YX\to X\). Its triviality says that the diagonal of \(f\) is an isomorphism, so \(f\) is \((-1)\)-truncated. If \(r\geq1\), then \(\Delta_f\) is \((r-1)\)-truncated and The induction hypothesis shows that \(\Delta_f\) is \((r-2)\)-truncated, and hence that \(f\) is \((r-1)\)-truncated.Now suppose that \(f\) is \(m\)-truncated for some \(m\) and that \(\pi_k(f)=*\) for every \(k>n\). If \(m\leq n\), there is nothing to prove. If \(m>n\), the sharper assertion successively lowers the truncation degree from \(m\) to \(m-1\), and eventually to \(n\).(2) Since all assertions are relative, we may work in the slice \(T_{/Y}\) and assume that \(f\) is the terminal map \(X\to *\). The case \(n=-1\) is Corollary 3.20.Suppose first that \(X\) is \(n\)-connected for \(n\geq0\). The map \(X\to *\) is left orthogonal to every \(n\)-truncated morphism. In particular, it is left orthogonal to every monomorphism, and is therefore an effective epimorphism. Since \(X\to *=\tau_nX\) is an \(n\)-truncation, Lemma 3.31 gives \(\pi_k(X)=*\) for every \(k\leq n\).Conversely, suppose that \(X\to *\) is an effective epimorphism and that \(\pi_k(X)=*\) for \(k\leq n\). Let \(t\colon X\to\tau_nX\) be the truncation map. It is \(n\)-connected by the factorization constructed in Proposition 3.18, hence it is an effective epimorphism. For \(k\leq n\), Lemma 3.31 gives Pullback along \(t\) is conservative by Lemma 2.29, so \(\pi_k(\tau_nX)=*\) for every \(k\leq n\). Since \(\tau_nX\) is \(n\)-truncated, the sharper assertion proved in part (1) successively lowers its truncation degree and shows that it is \((-1)\)-truncated. The composite is an effective epimorphism by assumption. Hence \(\tau_nX\to *\) is an effective epimorphism by Lemma 2.40. It is also a monomorphism, since \(\tau_nX\) is \((-1)\)-truncated, and is therefore an isomorphism. Thus \(X\) is \(n\)-connected.