base change for singular cohomology?

Question

Let $X,Y,Z$ be compact complex manifolds or smooth complex projective varieties, do we have the following commutative diagram similar to flat/proper/smooth base change for quasi-coherent sheaves? $$\require{AMScd} \begin{CD} H^\bullet(X\times Y,\mathbb{Z}) @>u^*>> H^\bullet(X\times Y\times Z,\mathbb{Z}) \\ @Vq_*VV @VVp_*V\\ H^\bullet(X ,\mathbb{Z}) @>v^*>> H^\bullet(X\times Z,\mathbb{Z}) \end{CD}$$ Here the maps are pullback/pushforward through either canonical projection or closed embedding.

Answer 1

This claim is weaker than base change.

By duality between homology and cohomology, it suffices to consider the dual diagram in homology, and check this commutes.

The dual to integration (which I guess is the pushforward map here) is taking the inverse image of homology cycles, and the dual to pullback is taking the image of homology cycles.

Then the claim is that if we take a cycle on $X \times Z$, the inverse image in $X \times Y$ of its image in $X$ is equal to the image in $X\times Y$ of its inverse image in $X\times Y \times Z$. But this is obvious.