Example 5.79.
For any topos \(T\), one has \(\Epi^{\,2} \subseteq \Conn_{\infty}\). Indeed, recall from Proposition 5.9 that every epimorphism is \(0\)-connected, so that
\[\Epi^{\,2} \subseteq \Conn_0^{\,2} = \Conn_2 \subseteq \Conn_1.\]
We claim that we also have an inclusion \(\Epi \cap \Conn_1 \subseteq \Conn_{\infty}\). Indeed, if \(f\colon X \to Y\) is an epimorphism, then we get a pushout square
If \(f \in \Conn_n\) for some \(n\), then \(\Delta_f \in \Conn_{n-1}\). The relative pushout product \(\Delta_f \ssquare_X \Delta_f\) is a base change of the ordinary pushout product, hence belongs to
\[\Conn_{n-1}\Conn_{n-1}=\Conn_{2n}.\]
Since the gap map of this square is \(f\), Blakers–Massey implies that \(f \in \Conn_{2n}\). Assuming \(f \in \Conn_1\), it follows inductively that \(f \in \Conn_{2^k}\) for all \(k\), so \(f \in \Conn_{\infty}\).