Let us consider an embedding of smooth manifolds $i: (Y, \partial Y) \rightarrow (X, \partial X)$. It is neat (see Hirsh, "Differential topology") if $i(\partial Y) = i(Y) \cap \partial X$ and, for every $y \in \partial Y$, there exists a boundary chart $(U, \varphi)$ of $X$ in $i(y)$ such that $U \cap i(Y)$ is the counterimage via $\varphi$ of the half-space orthogonal to the boundary. In the paper of Hopkins and Singer "Quadratic functions in geometry, topology and M-theory", appendix C.2, they define a generic map $f: (Y, \partial Y) \rightarrow (X, \partial X)$ to be neat if $f^{-1}(\partial X) = \partial Y$ and, for every $y \in \partial Y$, the map $df_{y}: T_{y}Y/T_{y}\partial Y \rightarrow T_{f(y)}X/T_{f(y)}\partial X$ is an isomorphism. Then they state the following theorem, whose proof I cannot find in the references they give:

Theorem C.17 If $f: Y \rightarrow X$ is a neat map of compact t-manifolds (for me it's enough to consider manifolds with boundary), there exist $N >> 0$ and a factorization $X \hookrightarrow Y \times \mathbb{R}^{N} \rightarrow Y$ of $f$ through a neat embedding $X \hookrightarrow Y \times \mathbb{R}^{n}$.

My questions are:

1) Is it true that the embedding $X \hookrightarrow Y \times \mathbb{R}^{n}$ is stably unique up to isotopy, as it happens for manifolds without boundary, and for embeddings $X \hookrightarrow \mathbb{R}^{n}_{+}$ of manifolds with boundary? Where can I find a proof?

2) Given a generic map $f: (Y, \partial Y) \rightarrow (X, \partial X)$ between compact smooth manifolds, is it always homotopic to a neat map, maybe realtively to $\partial Y$?


1 Answer 1


Regarding your question 2, yes all maps of pairs $(Y,\partial Y) \to (X, \partial X)$ are homotopic to neat maps. There are many ways to prove it but it boils down to a collar construction. I would be surprised if this (or something very similar to it) isn't in Hirsch, since he proves many similar things in that text. My mental picture of the construction is to imagine a wind blowing into $X$ from the boundary. With $Y$ "holding on" to $\partial X$ along $\partial Y$ but the rest being pushed in. But you can write down an explicit homotopy by using collars of $\partial X$ and $\partial Y$ in $X$ and $Y$ respectively.

For (1) the proof is the same as for embeddings into Euclidean space. Given two embeddings $X \to Y \times \mathbb R^n$ such that their projections $X \to Y$ agree, you can do the straight-line homotopy from one to the other. Provided $n$ is large enough, generically that homotopy can be made into an isotopy.

  • 2
    $\begingroup$ Nice mental picture! $\endgroup$
    – Mark Grant
    Feb 27, 2012 at 21:37

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