# Is $\pi_0$ of a homotopy cartesian square cartesian?

This is probably stupid but I don't see an argument.

Let \begin{eqnarray} P&\rightarrow & X\\ \downarrow&&\downarrow\\ Y&\rightarrow& Z \end{eqnarray} be a homotopy cartesian diagram of simplicial sets. You can assume for example that $X\to Z$ is a Kan fibration and the diagram is cartesian in the categorical sense.

Is \begin{eqnarray} \pi_0P&\rightarrow & \pi_0X\\ \downarrow&&\downarrow\\ \pi_0Y&\rightarrow& \pi_0Z \end{eqnarray} a cartesian diagram of sets?

-

## 1 Answer

This isn't true in spaces, certainly: take the homotopy pullback of $* \to S^1 \gets *$, which is $\Omega S^1$. We have that $*$ and $S^1$ are connected, but $\Omega S^1$ is not, since $\pi_0(\Omega S^1) = \pi_1(S^1) = \mathbb{Z}$.

Now hit everything with the singular complex functor.

-
As I feared, it was stupid. Thank you very much. I understand better now what I wanted. –  ferret Apr 18 '12 at 14:31