Let $f$ be transverse to $g$ so that $\require{AMScd}$ \begin{CD} V@>{f'}>> Z\\ @V{g'}VV @VV{g}V\\ Y @>{f}>> X \end{CD} is a Cartesian square of smooth manifolds. Then, if $g$ is proper and oriented, and $g'$ has the induced orientation, we get the formula

\begin{equation} \label{eq} f^*\circ g_* = g'_*\circ f'^* : H^*(Z;\mathbb R)\to H^{*+dim(X)-dim(Z)}(Y;\mathbb R). \end{equation}

There is nothing special in using real coefficients. Now, I'm going to follow de Rham and compute my real cohomology groups using currents. The space of currents on $M$, which I'll denote by $\mathcal D^*(M)$, is the space of continuous linear functionals on the space of smooth, compactly supported forms on $M$. We have the pushforward $$ g_*:\mathcal D^{*}(Z)\to \mathcal D^{*+dim(X)-dim(Z)}(X). $$ We cannot pull back any current along $f$. However, I believe since $f$ is transverse to $g$, $f^*$ is defined on $g_*$ of forms. I've so far failed to find a concrete reference for that statement, but I think it follows from [Theorem 8.2.4, 1]. Let me explain why I think that (as a response to Shifrins comment): The theorem says that $f^*T$ can be defined if $N(f)\cap WF(T)=\emptyset$. Here $N(f)$ is the ''set of normals" of $f$, defined by $$ N(f)=\{(x,\xi_x)\in X\times T^*X\backslash 0: \exists y\in Y\ s.t.\ \forall v\in T_yY \text{ we have } \xi(D_yfv)=0 \}. $$ The wave front set I do not have a good grasp of. But I think it is true that $WF(g_*\omega)\subset N(g)$ when $\omega$ is a smooth form. If that is true, one must only observe that when $f$ and $g$ are transverse, $N(f)\cap N(g)=\emptyset$. That's easy to see: If $(x,\xi)\in N(f)\cap N(g)$, we can choose $z\in g^{-1}(x)$ and $y\in f^{-1}(x)$ such that $\xi(D_yf T_yY+D_zg T_zZ)=0$. But then $\xi=0,$ since the sum we are applying it to is $T_xX$ since $f\pitchfork g$.

My main question is: does the equation $$ f^*\circ g_*=g'_*\circ f'^* $$ hold as maps $\Omega^*(Z)\to \mathcal D^{*+dim(X)-dim(Z)}(Y)$?

[1] Hörmander, Lars, The analysis of linear partial differential operators. I: Distribution theory and Fourier analysis., Moskva: ”Mir”. 464 p. (1986). ZBL0619.35001.

  • 1
    $\begingroup$ I don't understand how $\dim(Y)$ enters into $g_*$. The fact that $f\pitchfork g$ merely tells us that when $f(y)=g(z)=x$, $f_*T_yY + g_*T_zZ = T_xX$. Why is this relevant? Since $g$ pulls back $k$-forms on $X$ to $k$-forms on $Z$, shouldn't $g_*$ map $\mathcal D^{\dim Z-k}(Z)$ to $\mathcal D^{\dim X-k}(X)$, hence $\mathcal D^j(Z)$ to $\mathcal D^{\dim X-\dim Z+\ell}(X)$? $\endgroup$ Jan 27, 2022 at 19:20
  • $\begingroup$ Oops. $\ell$ should be $j$ in that final formula. $\endgroup$ Jan 27, 2022 at 19:41
  • $\begingroup$ If all we need is that the diagram commutes, I’m happy to assume just that! Regarding your comment about dim(Y), that was a typo. Thank you for pointing it out. I’m sorry I didn’t notice it prior to posting. $\endgroup$
    – Knaus
    Jan 27, 2022 at 23:12
  • $\begingroup$ Let me ponder the rest of the question. :) $\endgroup$ Jan 27, 2022 at 23:34
  • $\begingroup$ I have edited the question to include why I think transversality is relevant :) $\endgroup$
    – Knaus
    Jan 28, 2022 at 13:37


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