A short question on shriek maps
This should be easy but I don't quite see it. Let $M^m, N^n, X^d$ be compact, connected and oriented smooth manifolds. Let also $f:M\rightarrow X$ and $g:N\rightarrow X$ be transverse smooth maps. Then $$M\times_XN=\{(p,q)\in M\times N:\ f(p)=g(q)\}$$ is a smooth manifold. Consider $\pi_M,\pi_N$ the respective projections from it to $M,N$. The question is how to prove that $$(\pi_M)_*[M\times_XN]=f_!g_*[N]$$ as elements of $H_{m+n-d}(M;\mathbb{Z})$ (here $f_!$ is the shriek map $f_!=PD_M\circ f^*\circ PD_X^{-1}$).
N.B.: I can prove the result for real coefficients, but I would like to find a proof for integer coefficients.
Edit: In view of Stella's comment I'll write down the proof I have for real coefficients. Let me work in cohomology instead of homology: the equality I want to prove is equivalent to $$\pi_M^!(1)=f^*g^!(1).$$
Consider the projections $$\pi_1:X\times X\rightarrow X,\ \tilde{\pi}_M:M\times N\rightarrow M,$$ the inclusion map $$i:M\times_XN\rightarrow M\times N$$ and the map $$(f,g):M\times N\rightarrow X\times X.$$ Then we have (1) $\pi_1\circ(f,g)=f\circ\tilde{\pi}_M$, (2) $\pi_M=\tilde{\pi}_M\circ i$ and (3) $M\times_XN=(f,g)^{-1}(\Delta)$.
Moreover let $a_i$ be a basis for $H_{dR}(X)$ and let $b_i$ be its dual basis with respect to Poincaré duality. Then (4) $PD_{X\times X}^{-1}([\Delta])=\sum a_i\otimes b_i$.
Now, from (3) we have that $$i^!(1)=((f,g)^*\circ PD_{X\times X}^{-1})([\Delta]),$$ which from (4) leads to $$i^!(1)=\sum f^*(a_i)\otimes g^*(b_i).$$ We use (2) to get $$\pi_M^!(1)=(\tilde{\pi}_M^!\circ i^!)(1)=\tilde{\pi}_M^!\left(\sum f^*(a_i)\otimes g^*(b_i)\right).$$ Finally, since $\tilde{\pi}_M^!$ is integration along the fibre we get $$\pi_M^!(1)=\sum f^*(a_i)\int_Ng^*(b_i).$$
On the other hand, $g^!(1)=(PD_X^{-1}\circ g_*)[N]=\sum a_i\int_Ng^*(b_i)$, and applying $f^*$ we get the same expression as above.
I know the question is quite old, but maybe my sketch is of some use for future visitors.
What we want to proove: Every cartesian diagram of compact orientable manifolds $\require{AMScd}$ \begin{CD} A^a @>{p}>> M^m \\ @V{f}VV @VV{g}V\\ N^n @>{q}>>X^d \end{CD} (note $n+m = d + a$) induces a commutative diagram:
\begin{CD} H^*(A) @<{p^*}<< H^*(M) \\ @V{f_!}VV @VV{g_!}V\\ H^{*+n-a}(N) @<{q^*}<< H^{*+d-m}(X) \end{CD} There is an analogous version for homology. Now to prove this, instead of the Definition using Poincare duality, I recommend using the description via Thom-Isomorphisms described here (e.g.): https://ncatlab.org/nlab/show/fiber+integration
Inspired by this, first prove two special cases:
- $f$ and $g$ are both embeddings.
- $A$ and $M$ are both products and $f$ and $g$ are projection maps.
Now the general case follows by using Whitney to write $f$ (and $g$) as the composition of an emebedding followed by a projection, \begin{align*} A \overset{\iota}{\hookrightarrow} N \times S^{r} \overset{pr}{\twoheadrightarrow} N \end{align*} and using $f_! = pr_! \circ \iota_!$.