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Let $\pi \colon X \to B$ be a fibration with $B$ a path-connected CW complex. Write $B^p$ for the $p$-th skeleton of $B$ and set:

  • $X_p = \pi^{-1}(B^p)$,
  • $F_p^m = \ker [H^m(X) \to H^m(X_{p-1})]$, the kernel of a map induced by the inclusion $X_p \to X$.

A. Hatcher in his spectral sequences book writes on p.25 that:

The cup product in $H^∗(X; R)$ restricts to maps $F^m_p \times F^n_s \to F^{m+n}_{p+s}$.

The argument for that is that $F^m_p$ can be regarded as the image of the map $H^m(X, X_{p−1}) \to H^m(X)$ via the exact sequence of the pair $(X, X_{p−1})$, and then uses commutativity of a certain diagram (bottom of p. 26 in the mentioned book), part of which is a map $H^{m+n}(X\times X, X_p \times X \cup X \times X_s) \to H^{m+n}(X\times X, (X\times X)_{p+s})$. My question is: what sort of map is this? I I think it is induced by the inclusion $(X\times X, (X\times X)_{p+s}) \to (X\times X, X_p \times X \cup X \times X_s)$, and I think the proof requires this map to be a monomorphism (otherwise I don't see how to obtain commutativity of the diagram). But how to obtain the last statement?

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If it's commutativity of the diagram at the bottom of page 26 that you are worried about, this is just a formality. For convenience, insert another arrow into the diagram going from the middle group in the top row to the group in the lower right corner. This map is induced by inclusion (as is the map mentioned in your question). This map divides the diagram into a square and a triangle. The square commutes by naturality of relative cross product, and the triangle obviously commutes.

(This question probably belongs on math.stackexchange rather than mathoverflow.)

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