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Given a fibration $f:X \to B$ of CW complexes, it makes sense to guess that the pullbacks of a cell of $B$ will be a cell for $X$. That is, let $B_p$ be the $p$-th skeleton of $B$ and $X_p$ the pullback of $B_p \to B$ along $f$. Suppose also that $B_p$ is obtained from $B_{p-1}$ by attaching a single cell along $\varphi : S^{p-1} \to B_{p-1}$. Then we can define $\tilde{\varphi}:\tilde{S}^{p-1} \to X_{p-1}$ to be the pullback of $\varphi$ along $X_{p-1} \to B_{p-1}$. I'm trying to prove that $$X_{p} \simeq X_{p-1} \cup_{\tilde{\varphi}} \tilde{D^{p}}$$It seems to be a part in Hatcher's proof for the Serre spectral sequence (SSEQ thm 1.3). When I tried to prove it category-theoretically, I found out the claim I'm trying to prove is that if I have a map $T \to A\cup_C B$, then the incuded map $$(T\times _{A\cup_C B}A) \cup _{(T\times _{A\cup_C B}C)} (T\times _{A\cup_C B}B) \to T$$ is an equivalence. That is: A pullback of a pushout square along a map to the pushout is a pushout. Even though this is definitely the case in the category of sets, I can't see why this will be true in general, or at least for spaces.

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Well, apparently this is a special case of "colimits in spaces are universal": The general statement is that given a diagram $F:\mathcal{J}\to\mathcal{S}$ in the category of spaces, and a pair of maps $\operatorname{colim}\mathcal{J} \to Z$ and $Y\to Z$, there is an equivalence $$\left(\operatorname{colim}_\mathcal{J}F\left(i\right)\right)\times_Z Y \simeq \operatorname{colim}_\mathcal{J}\left(F\left(i\right)\times_Z Y\right)$$ The case here is the special case where the colimit is pushout and the map $\operatorname{colim}\mathcal{J} \to Z$ is the identity, and this is also known as Mather's second cube lemma, which was proven here at Theorem 25.

See also here for more information.

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