Homotopy colimit of a 3x3

Question

Hi I am wondering how you calculate homotopy colimits of a 3x3 diagram. In particular if we have (sorry not sure how to Tex these)

Top/bottom row: * <-- * --> *

Middle row: * <-- X --> *

With arrows mapping up from the middle row to the top and down from the middle row to the bottom.

I know the answer should be $\Sigma^2 X$ but I am not sure the process to calculate it.

Thanks!

Answer 1

You can compute a double homotopy colimit as an iterated homotopy colimit, that is, $\mathrm{hocolim}_{I \times J} F(i,j) \cong \mathrm{hocolim}_I \mathrm{hocolim}_J F(i,j)$. For your case, take the vertical homotopy colimits first, to get the diagram $\ast \leftarrow \Sigma X \to \ast$ (recall that for any $Y$ you have $\mathrm{hocolim}(\ast \leftarrow Y \to \ast) \cong \Sigma Y$ and that $\Sigma \ast \cong \ast$). Now take homotopy colimit of that to get $\Sigma^2 X$.