# Geometric intuition for Mather's cube theorem

Mather's cube theorem for the category of topological spaces says that given a homotopy-commutative cube:

If one pair of opposite faces are homotopy pushouts and the two remaining faces adjecent the source vertex are homotopy pullbacks, then the final two faces are also homotopy pullbacks.

What is the geometric intuition behind this theorem?

• Just to clear up my own confusion — this is Mather’s first cube theorem, correct? It seems some authors say “Mather’s cube theorem” to reference the second cube theorem. – Santana Afton Nov 2 '19 at 9:22
• @SantanaAfton, yes this is the first cube theorem. – Arrow Nov 2 '19 at 9:41

Say that $$B=B_1\cup B_2$$ and $$B_0=B_1\cap B_2$$. This is the second pushout square.
Now let $$E_1$$ be a bundle over $$B_1$$ and let $$E_2$$ be a bundle over $$B_2$$, and suppose that the restriction to $$B_0$$ is the same for both bundles -- call it $$E_0$$. These are the two given pullback squares. Now let $$E$$ be the union of $$E_1$$ and $$E_2$$ along $$E_0$$. This is the other pushout square. $$E$$ should be a bundle over $$B$$ whose restriction to $$B_1$$ (resp. $$B_2$$) is the bundle $$E_1$$ (resp. $$E_2$$).
• @Arrow, the theorem is about homotopy pushouts rather than pushouts, but I don't know how to give geometric intuition without switching to actual pushouts. And that means I should be looking at pushouts that are also homotopy pushouts, so at least one of the maps $B_0\to B_1$ and $B_0\to B_2$ should be some kind of nice injection. – Tom Goodwillie Nov 3 '19 at 16:26