Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
Let \begin{align} A &\to B\\ \downarrow &~~~~~\downarrow\\ C &\to D \end{align} be a homotopy pullback square of pointed simplicial sets. One gets a long exact sequence $$ \cdots\to\pi_n(A)\to \pi_n(B)\times \pi_n(C)\to \pi_n(D)\to \cdots\to\pi_0(A)\to \pi_0(B)\times \pi_0(C) $$ of homotopy groups as the homotopy pullback square defines a homotopy fiber sequence $\Omega D\to A\to B\times C$.
Does this sequence extend to the right by $\cdots\to \pi_0(D)$ or even by $\cdots\to \pi_0(D)\to 0$?
No, it doesn't -- in fact, there's not even a natural map! For instance, consider the case that $B=C=D=\mathrm{pt}_+$, with all maps the identity. The map in positive degrees uses the group structure on $\pi_{\geq 1}(D)$ in a crucial way.
