# Non-clean fiber products

Usually, the most general condition for fiber product of manifolds (or vector bundles) to exist is that we require the images cleanly intersects. See e.g. When do fibre products of smooth manifolds exist?

What happens when the intersection is not clean, say, of constant rank almost everywhere?

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In your last sentence, what do you mean by "of constant rank almost everywhere"? – Mark Grant Apr 2 '13 at 8:16
This question and answers might interest you: math.stackexchange.com/questions/322485/… – Mark Grant Apr 2 '13 at 8:16
The prototype example I have in mind is the following: Consider two bundle maps \begin{align*} \varphi_x: & T^*R^2 \rightarrow TR^2 \\ & dx \mapsto x\partial_y, dy \mapsto x\partial_x \\ \varphi_y: & T^*R^2 \rightarrow TR^2 \\ & dx \mapsto y\partial_y, dy \mapsto y\partial_x \end{align*} These two maps are transversal everywhere except at (0,0). In fact, they are not even clean at (0,0). But there is an "irreducible component" of the fiber product which is smooth. Question: How do we extract this smooth piece in general? – Songhao Li Jun 22 '13 at 18:51
The prototype example I have in mind is the following: Consider two bundle maps \begin{align*} \varphi_x: & T^*R^2 \rightarrow TR^2 \\ & dx \mapsto x\partial_y, dy \mapsto x\partial_x \\ \varphi_y: & T^*R^2 \rightarrow TR^2 \\ & dx \mapsto y\partial_y, dy \mapsto y\partial_x \end{align*} – Songhao Li Jun 22 '13 at 18:54
These two maps are transversal everywhere except at (0,0). In fact, they are not even clean at (0,0). But there is an "irreducible component" of the fiber product which is smooth, namely the vector bundle generated by $xy\partial_x$ and $xy\partial_y$. Question: How do we extract this smooth piece in general? – Songhao Li Jun 22 '13 at 18:55