Consider the Reedy category $2\rightarrow 1 \leftarrow 0$. Consider a map of diagrams of topological spaces $D\to E$ over this Reedy category:


![](https://i.stack.imgur.com/mNwpA.png)
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 ?
> 
> 
> 



About the dual of the cube lemma in homotopy theory
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.)


I apologise for the lack of diagrams. I think it is easier to understand if you draw them yourself.


