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?

$\begingroup$ 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? $\endgroup$ – Brian Lins Jul 18 '14 at 23:40

2$\begingroup$ 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? $\endgroup$ – 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 $\mathbb{R}^n$.
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 PontryaginThom construction). However, it is much stronger than that. These kinds of transverse homotopies have been studied by Jon Woolf and his collaborators (see http://uk.arxiv.org/abs/0910.3322, 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 http://uk.arxiv.org/abs/0910.3322 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 http://pcwww.liv.ac.uk/~jonwoolf/Slides/belfasttalk.pdf might be helpful.

$\begingroup$ 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. $\endgroup$ – Mark Grant Jul 18 '14 at 16:35