Theorem 3.22.
Let \(n \geq 0\). Then a morphism \(f\colon X \to Y\) is \(n\)-connected if and only if it is an effective epimorphism and \(\Delta_f\colon X \to X \times_Y X\) is \((n-1)\)-connected.
Proof
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
\[\pi_0(f)=* \quad\Longleftrightarrow\quad \Delta_f\text{ is an effective epimorphism}.\]
Combining this equivalence with the higher homotopy-group conditions proves the theorem.