# About the dual of the cube lemma in homotopy theory

Consider the Reedy category $$2\rightarrow 1 \leftarrow 0$$. Consider a map of diagrams of topological spaces $$D\to E$$ over this Reedy category: The maps which are fibrations are depicted with the symbol $$\twoheadrightarrow$$: the map of diagrams $$D\to E$$ is a pointwise fibration, and moreover the maps $$D_2\to D_1$$ and $$E_2\to E_1$$ are fibrations as well. This implies that the two diagrams $$D$$ and $$E$$ are Reedy fibrant. We suppose moreover that the map $$D_2 \twoheadrightarrow E_2 \times_{E_1} D_1$$ (not depicted in the image) is a fibration as well. Therefore the map of diagrams $$D\to E$$ is a Reedy fibration. For the Reedy model structure, the inverse limit is a right Quillen functor. This implies that the map $$f$$ from the inverse limit of D to the inverse limit of E is a fibration.

Here is now the question.

With these hypotheses, is it sufficient to conclude that

$$\varprojlim D \longrightarrow \varprojlim E \times_{E_0} D_0$$

is a fibration as well ? If not, what is missing ?

Yes, $$D_0 \times_{D_1} D_2 \to (E_2 \times_{E_1} E_0) \times_{E_0} D_0$$ is a fibration.
First, observe that $$(E_2 \times_{E_1} E_0) \times_{E_0} D_0 \cong E_2 \times_{E_1} D_0 \cong (E_2 \times_{E_1} D_1) \times_{D_1} D_0$$ by the pullback pasting lemma. Also, $$D_2 \times_{D_1} D_0 \cong D_2 \times_{E_2 \times_{E_1} D_1} ((E_2 \times_{E_1} D_1) \times_{D_1} D_0)$$ but we assumed that $$D_2 \to E_2 \times_{E_1} D_1$$ is a fibration, so $$D_2 \times_{E_2 \times_{E_1} D_1} ((E_2 \times_{E_1} D_1) \times_{D_1} D_0) \to (E_2 \times_{E_1} D_1) \times_{D_1} D_0$$ is also a fibration. But this is isomorphic to $$D_0 \times_{D_1} D_2 \to (E_2 \times_{E_1} E_0) \times_{E_0} D_0$$ so we are done. (So, in fact, we don't need to assume anything else is a fibration.)