# Multiple Transversal Pullback

Suppose we have three smooth manifolds $M_1$, $M_2$ and $N$ and two smooth maps $f_1:M_1 \rightarrow N$ and $f_2:M_2 \rightarrow N$. Than an important and central construction in differential topology is the $transversal$ $pullback$

$$M_1 \times_{f_1Nf_2} M_2 = \\lbrace\left(x_1,x_2 \right) \in M_1 \times M_2 |f_1(x_1)=f_2(x_2) \rbrace$$

A proof that it is a manifold goes like:

$M_1 \times_{f_1Nf_2} M_2 = (f_1 \times f_2 )^{−1}(\Delta)$, where $f_1 \times f_2 : M_1 \times M_2 \rightarrow N \times N$ and where $\Delta$ is the diagonal of $N \times N$ , and $f_1 \times f_2$ is transversal to $\Delta$ if and only if $f_1$ and $f_2$ are transversal.

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Now the question is, can we extend this to multiple transversal pullbacks? For example a "three times pullback":

$M_1 \times_{f_1Nf_2} M_2 \times_{f_2Nf_3} M_3 = \lbrace \left(x_1,x_2,x_3 \right) \in M_1 \times M_2 \times M_3 |f_1(x_1)=f_2(x_2); f_2(x_2)=f_3(x_3) \rbrace$

is this well defined as a smooth manifold and if yes how is it proofed?

And is there another generalization to the $n$-times transversal pullback?

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Nothing at all ? – Mirco Dec 26 '11 at 14:01
You could define $f_1: M_1 \to Z, f_2: M_2 \to Z, f_3: M_3 \to Z$ to be transversal if $T_zZ \cong \Im(df_1) + \Im(df_2) + \Im(df_3)$, for all $x \in M_1, y \in M_2, w \in M_3$ such that $f_1(x)=f_2(y)=f_3(w)=z$. The obvious definition of transversal pullback would be as the limit of the diagram with legs $f_i$, that is, a manifold $M$ with smooth $\pi_i: M \to M_i$ such that $f_1\pi_1=f_2\pi_2=f_3\pi_3$, and universal with this property, i.e if $(N,\phi_1, \phi_2, \phi_3)$ satisfy similar identities, then there is a unique smooth map $\psi: N \to M$ such that $\pi_i\psi=\phi_i$, ($i=1,2,3$). – José Siqueira Mar 5 at 10:42