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I'll begin with the question, which is intrinsically interesting:

Let M be a manifold with some submanifold Y. Suppose that $W \rightarrow M$ is a smooth, proper map. Does there exist another map $W \rightarrow M$ homotopic to the original that is ALSO proper and transverse to the submanifold Y?

Let me note that I am definitely not assuming the manifolds are compact.

Why I care: This question came up while thinking about the geometric description of complex cobordism given by Quillen in "Elementary proofs of some results of cobordism theory using Steenrod operations." I have a geometric description of the coboundary map in the Mayer-Vietoris sequence but as of now it relies on the answer to the above question being "yes."

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    $\begingroup$ Yes, there are such maps. You obtain them in the standard way -- via small perturbations. You can be careful and make the perturbations on a collection of compact subspaces of $W$ that exhaust it. But it's the standard transversality argument. It appears in Guillemin and Pollack, for example look at their proof of the Whitney theorem in the case the manifold is non-compact. $\endgroup$
    – Ryan Budney
    Commented Nov 5, 2012 at 21:22
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    $\begingroup$ @Dylan: Put a complete Riemannian metric on $M$ and consider a deformation of your map $W\to M$ by a generic vector field which vanishes at infinity in the sense that the norm $|V(x)|$ tends to zero as the distance $d(x,x_0)$ diverges to infinity. This ensures preperness of the deformation. Transversality follows from the same argument as in the compact case (Sard's theorem). $\endgroup$
    – Misha
    Commented Nov 5, 2012 at 23:23
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    $\begingroup$ Another way to say Misha's argument is that the property of being proper is an open condition provided you're using the uniform topology on your function space, but you have to use a complete Riemann metric to generate the uniform topology. Guillemin and Pollack avoid that since they're exhausting the manifold by compact sets -- in a sense they're just avoiding all that formalism with a short-cut to the result. $\endgroup$
    – Ryan Budney
    Commented Nov 6, 2012 at 0:14
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    $\begingroup$ Or stated another way, I think you're trying to think about this too categorically Dylan. The construction is very direct. $\endgroup$
    – Ryan Budney
    Commented Nov 6, 2012 at 0:35
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    $\begingroup$ Thank you both, that clears things up! And yes, I agree I am out of touch with my concrete side :) $\endgroup$
    – Dylan Wilson
    Commented Nov 6, 2012 at 1:44

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Regarding the coboundary map in complex cobordism, I think that this was done by Dold in "Geometric Cobordism and the Fixed Point Transfer" (at least for oriented cobordism) in 2.10.

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  • $\begingroup$ !! Thank you thank you! This is exactly what I was looking for. I'm going to accept this answer out of sheer gratitude, but any wandering MOers should be aware that the answer to the question in the title can be found in the comments and in Sergei's lovely answer. Thanks again Haggai! $\endgroup$
    – Dylan Wilson
    Commented Aug 19, 2013 at 18:33
  • $\begingroup$ I'm glad it helps you. You can also look at Kreck's book "differential algebraic topology" where he does it to a "stratifold cohomology" which is a bordism approach to integral cohomology. There it is done very carefully (appendix B). him.uni-bonn.de/fileadmin/user_upload/kreck-DA.pdf $\endgroup$
    – Haggai Tene
    Commented Aug 20, 2013 at 7:23
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The comments explain how to prove the fact.

If you want to put a formal wrapping around it, consider the strong (Whitney) $C^\infty$ topology on the space of maps $W\to M$. The strong $C^0$ topology on $C(M,\mathbb R)$ can be defined as follows: for every function $h:M\to\mathbb R$ which is positive and locally bounded away from 0 (but may tend to 0 at infinity), declare the set $$U_h:=\{f\in C^0(M): \forall x\in M \ |f(x)|<h(x)\}$$ a neighborhood of zero; this gives you a prebase of the topology. For smooth maps between manifolds the definition is similar but involves derivatives and a locally finite covering by charts (or, alternatively, a complete Riemannian metric, on which the resulting topology does not depend).

In the strong $C^\infty$ topology, the set of proper maps is open, and the set of maps transverse to $Y$ is open and dense. For a reference, see e.g. Hirsch, "Differential topology" (1976), Chapter 2 and Theorem 2.1(a) in Chapter 3.

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