pullback and fiber sequence

Question

Let $A\rightarrow D\leftarrow C$ a diagram of connected pointed toplogical space where $A\rightarrow D$ is a fibration. Denote $P=A\times_{D}C$. We obtain a homotopy fiber sequence $$ \Omega D\rightarrow P\rightarrow A\times C $$

If we suppose that $D=\Omega X$ for some pointed topological space $X$. Do we obtain a homotopy fiber sequence $$ P\rightarrow A\times C\rightarrow D ?$$ where the map $A\times C\rightarrow D$ is obtained as a composition $A\times C\rightarrow D\times D\rightarrow D$ (the second map is a concatenation of loops)

Answer 1

Yes. Here are some details.

1. The space $P$ sits in homotopy pullback diagram $\require{AMScd}$ $$ \begin{CD} P @>>> D \\ @VVV@VVV \\ A\times C @>>> D\times D \end{CD} $$ where the the right vertical map is the diagonal. In fact, one can see this by replacing the latter map with the free path fibration $D^I \to D\times D$. After this replacement the diagram becomes a pullback and a pullback with one of the terminal maps a fibration is always a homotopy pullback.

2. When $D = \Omega X$, there the diagonal is induced by the map $m:D\times D \to D$ given by $(\gamma_1,\gamma_2) \mapsto \gamma_1 \cdot \bar\gamma_2$, where the bar means loop inversion. This means that there is a commutative homotopy pullback $$ \begin{CD} D @>\text{diag} >> D\times D \\ @VVV @VV m V \\ * @>>> D \end{CD} $$ where $*$ is some contractible space. (Perhaps the easiest way to see this is to note that the diagonal $X\to X \times X$ has homotopy fiber $D$.)

3. The base change of a map which is induced in also induced: this means that there is a commutative homotopy pullback $$ \begin{CD} P @>>> A \times C \\ @VVV @VVV \\ * @>>> D \end{CD} $$ where the bottom map is the same as in the previous diagram.