Homotopy group objects

Since \(F\colon T_{/X} \to T'_{/F(X)}\) preserves finite limits and \(X^{S^n}\) is a finite limit, we see that \(F(X^{S^n}) \cong F(X)^{S^n}\). Moreover, a colimit-preserving functor between presentable categories is a left adjoint. Thus \(F\) admits a right adjoint, and Lemma 3.8 shows that it commutes with \(\tau_0\).

Cotensoring by \(S^n\) preserves products, and Lemma 3.9 shows that \(0\)-truncation preserves products in every slice topos. Applying the definition therefore gives

When \(T = \An\), this is the usual long exact sequence of a homotopy fiber, including its group action on \(\pi_0\) in the final nonabelian degree. The claim follows pointwise for any presheaf topos \(\PSh(C)=\Fun(C\catop,\An)\). By Theorem 2.42, an arbitrary topos is a left exact localization of a presheaf topos. The localization preserves the finite limits, colimits, and homotopy group objects used in the construction, by Lemma 3.25, so it carries the pointwise long exact sequence to the asserted sequence in \(T\). This is the argument of [Lurie 2009, Remark 6.5.1.5].

Choose a left exact localization \(L\colon \PSh(C)\to T\) with fully faithful right adjoint \(R\). Form the \(n\)-truncation \(R(X)\to\tau_n^{\mathrm{pre}}R(X)\) in \(\PSh(C)\). Truncations and homotopy group objects in the presheaf topos are computed pointwise, so classical homotopy theory gives isomorphisms on \(\pi_k\) for \(k\leq n\). Applying \(L\) preserves these homotopy group objects by Lemma 3.25. It also identifies \(L\tau_n^{\mathrm{pre}}R(X)\) with \(\tau_nX\cong Y\) by Lemma 3.8. The resulting isomorphisms are precisely those in the statement. See also [Lurie 2009, Lemma 6.5.1.9].

(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.

We must show that \(f\) is \(n\)-connected if and only if it is an effective epimorphism and \(\Delta_f\) is \((n-1)\)-connected.By the proposition, \(f\) is \(n\)-connected if and only if it is an effective epimorphism and \(\pi_k(f)=*\) for \(0\leq k\leq n\). Similarly, \(\Delta_f\) is \((n-1)\)-connected if and only if it is an effective epimorphism and \(\pi_k(\Delta_f)=*\) for \(0\leq k\leq n-1\). By Remark 3.28, the latter vanishing condition is equivalent to \(\pi_k(f)=*\) for \(1\leq k\leq n\). It remains to compare the two conditions in degree zero.Work in the slice topos \(T_{/Y}\), write \(Z\) for the object corresponding to \(f\), and let \(U=\tau_0Z\). Suppose that \(f\), or equivalently \(Z\to *\), is an effective epimorphism. Then \(U\to *\) is an effective epimorphism by Lemma 3.13. By the description of \(\pi_0\) in Example 3.24, the morphism \(\pi_0(f)\to Z\) is the pullback of \(U\to *\) along \(Z\to *\). Since pullback along the effective epimorphism \(Z\to *\) is conservative by Lemma 2.29, we have \(\pi_0(f)=*\) if and only if \(U\to *\) is an isomorphism.On the other hand, Lemma 3.13 and Lemma 3.9 show that \(\Delta_f\) is an effective epimorphism if and only if the diagonal \(\Delta_U\colon U\to U\times U\) is an effective epimorphism. Since \(U\) is \(0\)-truncated, \(\Delta_U\) is a monomorphism. Consequently, if \(\Delta_U\) is an effective epimorphism, it is an isomorphism, so \(U\to *\) is a monomorphism. Together with the effective epimorphism \(U\to *\), this makes \(U\) terminal. The converse is immediate. We have therefore proved, under the effective-epimorphism hypothesis on \(f\), that Combining this equivalence with the higher homotopy-group conditions proves the theorem.

Dual to the proof of Lemma 3.7(2), we may factor \(f\) as \(X \xrightarrow{(\id_X,f)} X \times_Z Y \xrightarrow{\pr_Y} Y\), where the first map is a base change of \(\Delta_g\), hence \(n\)-connected by the theorem, and the second map is a base change of \(gf\).

By the theorem, \(X\) is \(n\)-connected if and only if \(\Delta\colon X \to X \times X\) is \((n-1)\)-connected. Given a base point \(x\colon * \to X\), it follows from \(0\)-connectedness and Corollary 3.33 that \(x\) is \((-1)\)-connected. Hence \((x,x)\colon * \to X \times X\) is an effective epimorphism. By Lemma 3.19, the diagonal \(\Delta\) is \((n-1)\)-connected if and only if its base change \(\Omega X \to *\) is \((n-1)\)-connected.For truncatedness, \(X\) is \(n\)-truncated if and only if \(\Delta\) is \((n-1)\)-truncated. The same locality lemma identifies this with \((n-1)\)-truncatedness of \(\Omega X\to *\).

Full faithfulness of \(f^*\) was proved in Proposition 3.18.For essential surjectivity, we may replace \(T\) by \(T_{/Y}\) and assume \(Y = *\). Given an \((n-1)\)-truncated morphism \(U \to X\), we must show that it is in the image of the functor \((-) \times X\colon T_{\leq n-1} \to (T_{/X})_{\leq n-1}\). We claim that the map \(U \to \tau_{n-1} U \times X\) is an isomorphism. By the factorization system from Proposition 3.18, it suffices to show it is both \((n-1)\)-truncated and \((n-1)\)-connected.

• The map is \((n-1)\)-truncated since it lives over \(X\), and both maps to \(X\) are \((n-1)\)-truncated. The claim follows by left cancellation.
• To see it is \((n-1)\)-connected, note that it factors as
  \[U \longrightarrow U \times X \longrightarrow \tau_{n-1} U \times X.\]
  The first map is \((n-1)\)-connected, as it is a base change of the diagonal \(X \to X \times X\), which is \((n-1)\)-connected by the theorem. The second map is \((n-1)\)-connected since it is a base change of the map \(U \to \tau_{n-1} U\).