If {M} is a manifold and {N} a compact submanifold, then a tubular neighborhood of {N} consists of an open set {U \supset N} diffeomorphic to a neighborhood of the zero section in some vector bundle {E} over {N}, by which N corresponds to the zero section.

Theorem 1 Hypotheses as above, {N} has a tubular neighborhood.

When {M= \mathbb{R}^n}, the idea is to take {E} as the normal bundle of {N}, whose fiber at {p} consists of tangent vectors in {\mathbb{R}^n} perpendicular to {T_p(N)}. There is a map {f: E \rightarrow \mathbb{R}^n} sending {(v,p)} for {v \in E_p, p \in N} to {v+p \in \mathbb{R}^n}. When restricted to the zero section, this map is just the identity on {N}, and the differential is the natural map

\displaystyle E_p \oplus T_p(M) \rightarrow \mathbb{R}^n, 

which is an isomorphism, by construction. Using the inverse function theorem, we can find a bounded neighborhood {V \subset E} containing the zero section, such that {f: V \rightarrow \mathbb{R}^n} is locally a homeomorphism.

Now, following Spivak, I quote a lemma:

Lemma 2 If {X' \subset X} is a closed subset of the compact metric space {X}, and {f: X \rightarrow Y} is a local homeomorphism that is injective on {X'}, then there is a neighborhood {U \supset X'} with {f: U \rightarrow Y} injective.

Suppose this failed; then we can find a sequence {U_n} of open sets containing {X'} (e.g. the {\frac{1}{n}} neighborhood) with {\bigcap U_n = X'} such that {f} isn’t one-to-one on {U_n}. Thus one picks {x_n,y_n \in U_n} with {f(x_n)=f(y_n), x_n \neq y_n}. By taking subsequences if necessary, assume both {x_n,y_n} converge to {x,y \in X'}. Then clearly {f(x)=f(y)}. I claim {x \neq y}; otherwise for {n} very large there would be {x_n \neq y_n } close to {x} with {f(x_n)=f(y_n)}, and this contradicts the local homeomorphism condition. Now {x \neq y} but {f(x)=f(y)}, so this contradicts the injectivity on {X'}.

So returning to the special case of the theorem, we can find a suitably small neighborhood {U \subset E} such that the map {f: U \rightarrow \mathbb{R}^n} is injective, hence an isomorphism onto the image.

Now for the general case. We can make {M} into a metric space via a Riemannian metric, which can be constructed through a standard partition-of-unity argument. Similarly, we can obtain a connection on {M} (not necessary for it to be compatible with the Riemannian metric), which leads to exponential maps as before.

We can restrict the bundle {TM} to {N} to get a vector bundle {TM|N}, which contains as a subbundle {TN}. Let {E} be a complement (e.g. obtained by putting a nondegenerate inner product on {TM|N}) of {TN \subset TM|N}, i.e. such that {E \oplus TN \simeq TM|N}.

Now there is an open neighborhood {U} of the zero section in {E} and the exponential map {U \rightarrow M}; this is smooth by the ODE theorem, and the differential at {0_p \in E_p} for {p \in N} is seen to be an isomorphism as before (yesterday we computed the differential of the exponential map at {0}). This fact, with the lemma above, completes the proof of the tubular neighborhood theorem.

Note in particular that {N} is a smooth deformation retract of the tubular neighborhood {U} (via {v \rightarrow (1-t)v}, {0\leq t \leq 1}).

Here is another variant of this result:

Theorem 3 (Collar Neighborhood Theorem) Let {N} be a manifold-with-boundary, with compact boundary {\partial N}. There is a neighborhood {U \supset \partial N} isomorphic to {\partial N \times I} for {I \subset \mathbb{R}} an open interval.

Choose a Riemannian metric on {N} (which is defined in the same way as for a regular manifold). On each point {p} of the boundary {\partial N}, we can assign a normal {n_p \in T_p(N)} perpendicular to {T_p(\partial N)} that points “inward” to {N}. We can find a connection {\nabla} on {N}, and using the exponential map, we can consider {F(p,t) = \exp_p( t n_p)}. Then {F} is defined on {\partial N \times I} for {I} small enough. Now this is a local isomorphism at each point of {\partial N \times 0}, and we can apply the same type of argument as before.