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Let $M$ and $N$ be submanifolds of $\mathbb{R}^n$ and let $a(t)$ be a smooth path in $\mathbb{R}^n$ such that $M+a(t)$ intersects $N$ transversally for all $t \in [0,1]$. Is there a nice relationship between the submanifolds $(M+a(0)) \cap N$ and $(M+a(1)) \cap N$? Are they homotopy equivalent or better yet homeomorphic?

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Thank you Oscar and Mark. Your answers are both helpful. This seems like a natural question, and given that the answer is that the two submanifolds are homeomorphic (even diffeomorphic), this result must be well known. I wonder if there is a reference where it is stated explicitly? – Brian Lins Jul 18 '14 at 23:40
For the answer below: Don't we need some compactness assumptions on the submanifolds? What about $(0,1)\times\{0\}$ and $\{0\}\times (-1,1)$ where the map is given by a shift? – Thomas Rot Jul 19 '14 at 10:15

Let me rephrase the construction: you have a map $$\varphi : [0,1] \times M \to \mathbb{R}^n$$ which for each $t \in [0,1]$ is an embedding ($\varphi(t,x) = x+a(t)$ in your notation) and is transverse to $N \subset \mathbb{R}^n$. In particular, it follows that the map $\varphi$ is transverse to $N$, and so $$X:=\varphi^{-1}(N) \subset [0,1] \times M$$ is a submanifold. If we write $X_t := \varphi(t,-)^{-1}(N)$ then this is a smooth manifold for all $t$, as $\varphi(t,-)$ is transverse to $N$: thus $X$ is a cobordism from $X_0$ to $X_1$, and these are the two manifolds in your question.

I claim that the projection map $\pi : X \to [0,1]$ is a submersion. To see this, note that for $(t,m) \in X$ the map $$T_{(t,m)}([0,1] \times M) \to T_t[0,1]$$ is surjective, so we can find a $v \in T_{(t,m)}([0,1] \times M)$ mapping to $d/dt$. Under the surjective map $$T_{(t,m)}([0,1] \times M) \overset{D\varphi}\to T_{\varphi(t,m)}\mathbb{R}^n \to \nu_{\varphi(t,m)}N$$ to the normal bundle of $N$, $v$ may not map to 0, but as precomposing with $$T_{(t,m)}(X_t) \to T_{(t,m)}([0,1] \times M)$$ the map stays surjective, we can modify $v$ by an element of $T_{(t,m)}(X_t)$ (which does not change its projection to $T_t[0,1]$) to get a new $v'$ mapping to 0 in $\nu_{\varphi(t,m)}N$; hence $v' \in T_{(t,m)}(X)$, and it maps to $d/dt$ under $D\pi$, as required.

By picking a splitting of the bundle epimorphism $$TX \to \pi^*T[0,1]$$ we get a vector field on $X$, and integrating this shows that $X \cong [0,1] \times X_0$. Thus in particular $X_1$ and $X_0$ are diffeomorphic, but even better, $\varphi\vert_X$ gives an isotopy between them in $\mathbb{R}^n$.

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I believe they are homeomorphic.

Let $j: M\hookrightarrow\mathbb{R}^n$ denote the embedding of $M$. Your path defines a smooth homotopy $h: M\times [0,1]\to \mathbb{R}^n$ given by $h(x,t) = j(x)+a(t)$, and the assumption is that each $h_t = h(-,t): M\to \mathbb{R}^n$ is transverse to $N\subseteq \mathbb{R}^n$.

This assumption implies firstly that $h$ is transverse to $N$, and therefore that $W:=h^{-1}(N)\subseteq M\times [0,1]$ is a cobordism between $h_0^{-1}(N)=(M+a(0))\cap N$ and $h_1^{-1}(N)=(M+a(1))\cap N$ (cf. the Pontryagin-Thom construction). However, it is much stronger than that. These kinds of transverse homotopies have been studied by Jon Woolf and his collaborators (see, for example). The key observation is that, under these conditions, the map $W\subseteq M\times [0,1]\to [0,1]$ which projects onto the interval has no critical points (see Theorem 2.3, which is about maps transversal to stratifications but the same reasoning applies to maps transversal to submanifolds). Therefore $W$ is a trivial cobordism, and the two ends are homeomorphic (this uses the ideas of Morse theory).

I guess this could be better explained. The slides here might be helpful.

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Oscar beat me to it by a few minutes, but I thought I'd post the answer anyway in case you find the references useful. – Mark Grant Jul 18 '14 at 16:35

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