Connectivity

(1) We first show that any map \(f\colon X \to Y\) factors as an \(n\)-connected map followed by an \(n\)-truncated map. Let \(X_n := \tau_n(X/Y)\) be the \(n\)-truncation of \(X \in T_{/Y}\). The map \(X_n \to Y\) is \(n\)-truncated by definition.To show that \(X \to X_n\) is \(n\)-connected, consider any \(n\)-truncated morphism \(U \to X_n\). We must show that the map is an equivalence. Since \(T_{/X_n} \simeq (T_{/Y})_{/X_n}\), this map is induced on vertical fibers in the following diagram of Hom animae in \(T_{/Y}\):

Since \(U\) and \(X_n\) are \(n\)-truncated objects in \(T_{/Y}\), the universal property of \(X_n = \tau_n(X/Y) \in T_{/Y}\) guarantees that both horizontal maps are equivalences. Thus the induced map on fibers is also an equivalence.Next, we show that any \(n\)-connected map \(f\colon X \to Y\) is left orthogonal to any \(n\)-truncated map \(g\colon W \to Z\). We must show that the anima of diagonal fillers in any commutative square

is contractible. Since \(n\)-truncated maps are closed under base change, we may replace \(g\) by the projection \(W \times_Z Y \to Y\). This reduces the problem to finding diagonal fillers in the commutative square:

The anima of such diagonal fillers is contractible by \(n\)-connectedness of \(f\). We have therefore constructed the required factorization and proved orthogonality, which are exactly the two conditions in Definition A.2.(2) Consider a pullback square

By Corollary 3.10, we have \(\tau_n(X'/Y') \cong g^*\tau_n(X/Y)\). If \(f\) is \(n\)-connected, then \(\tau_n(X/Y) = Y\), so \(\tau_n(X'/Y') = Y'\). Hence \(f'\) is also \(n\)-connected.(3) We need to show that \(f\) is \(n\)-connected if and only if for each two \(n\)-truncated maps \(Z \to Y\) and \(Z' \to Y\), the induced map is an equivalence. By the universal property of \(X \times_Y Z'\), the latter is equivalent to the condition that composition with the projection \(X \times_Y Z \to Z\) induces an equivalence In other words, we have to show that \(f\) is \(n\)-connected if and only if the map \(X \times_Y Z \to Z\) is left orthogonal to every \(n\)-truncated map \(Z' \to Y\). If \(f\) is \(n\)-connected, this follows from parts (1) and (2), since \(X \times_Y Z \to Z\) is a base change of \(f\). Conversely, take \(Z=Y\). The resulting orthogonality condition says that \(f\) is left orthogonal to every \(n\)-truncated map over \(Y\), which is equivalent to \(n\)-connectedness by the characterization above.

Both properties are preserved by arbitrary base change, by Lemma 3.5 and Proposition 3.18. We prove the converses.For truncatedness, argue by induction on \(n\). If \(n=-2\), the claim says that an isomorphism can be detected after pullback along an effective epimorphism, which follows from Lemma 2.29. For \(n\geq -1\), the diagonal of \(f'\) is obtained from the diagonal of \(f\) by base change along the effective epimorphism The induction hypothesis therefore shows that \(\Delta_f\) is \((n-1)\)-truncated whenever \(\Delta_{f'}\) is.For connectedness, Corollary 3.10 gives an isomorphism If \(f'\) is \(n\)-connected, the pullback of \(\tau_n(X/Y)\to Y\) along \(g\) is therefore an isomorphism. Pullback along \(g\) is conservative by Lemma 2.29, so \(\tau_n(X/Y)\to Y\) is itself an isomorphism. Thus \(f\) is \(n\)-connected.

Both the \((-1)\)-connected maps as well as the effective epimorphisms are characterized as the morphisms that are left orthogonal to the monomorphisms/(-1)-truncated maps.

A basepoint \(x\colon * \to X\) determines a section of the map \(X \to *\), which by Lemma 2.38 implies that \(X \to *\) is an effective epimorphism, hence \((-1)\)-connected.