In the finite-dimensional setting, it's possible to construct tubular neighborhoods without anything like Godement's lemma. Many sources simply rely on a point-set topology argument that's based on the same idea as Godement's lemma (to be precise, I'm talking about the argument on p. 109 of Lang's book Differential and Riemannian Manifolds, which he says follows Godement). I'll explain another approach.

The idea is to use a Riemannian metric on the manifold $M$, which also induces a Riemannian metric on $TM$ (viewed as a manifold in its own right). The geodesic distance then gives a (topological) metric on $TM$. If $Y\subset M$ is a (not necessarily closed) submanifold, then a simple metric geometry argument can then be used to find a neighborhood of the zero section of $N(Y)$ (thought of as the perpendicular complement of $TY$ inside $TM$) on which the exponential map is injective. The key fact about the exponential map $f$ is that every point in the zero section of $N(Y)$ has a neighborhood on which $f$ is a diffeomorphism onto an open subset of $M$. (Edit: Note that in the finite dimensional setting, $N(Y)$ is automatically a locally trivial vector bundle. This does not seem to be the case for arbitrary infinite dimensional Riemannian manifolds, as discussed here: Orthogonal complements in Hilbert bundles. Hence the discussion that follows does not work in as great generality as arguments based on Godement's lemma.)

The general metric geometry fact is this:

Consider a metric space $T$ and subspaces $X, Y$, and $D$ such that $Y \subset X$ and $Y\subset D$. (Think: $T = TM$, $X$ is the zero section of $TM$, $Y$ is a submanifold of $M$, and $D$ is the domain of the exponential map, lying inside $NY$.) Let $f: D\to X$ be a continuous map that restricts to the identity on $Y$ (think: $f$ is the exponential map). Assume further that for each $y\in Y$ there exists $\epsilon_y >0$ such that $f$ restricted to $B_{\epsilon (y)} (y, D) = \{z\in D \,:\, d(z,y) < \epsilon(y)\}$ is a homeomorphism onto an open subset of $X$. Then there exists a subspace $D'$, open in $D$, on which $f$ is injective.

Proof. For each $y\in Y$, $f(B_{\epsilon (y)/2} (y, D))$ is open in $X$, hence contains $B_{\epsilon'(y)} (y, X)$, for some $\epsilon'_y < \epsilon_y/4$ (remember that $f(y) = y$). Now consider the inverse image $Z_y$ of $B_{\epsilon'_y} (y, X)$ under the restriction of $f$ to
$B_{\epsilon (y)/2} (y, D)$. Since $f$ is a homeomorphism when restricted to this ball, $Z_y$ is open as a subset of $D$. Now I claim that $f$ is injective on $D' = \bigcup_{y\in Y} Z_y$. Say $f(z_1) = f(z_2) = y_0$ with $z_1 \in Z_{y_1}$ and $z_2\in Z_{y_2}$, and assume $\epsilon_{y_1} \geq \epsilon_{y_2}$. Then we have
$$d(z_2, y_1) \leq d(z_2, y_2) + d(y_2, y_0) + d(y_0, y_1) < \epsilon_{y_2}/2 + \epsilon'_{y_2} + \epsilon'_{y_1} $$
$$< \epsilon_{y_1}/2 + \epsilon_{y_2}/4 + \epsilon_{y_1}/4 \leq \epsilon_{y_1}$$
(for the second inequality, note that by definition, $y_0 = f(z_i) \in f(Z_{y_i}) \subset B_{\epsilon'_{y_i}} (y_i, X)$ for $i=1, 2$).
So $z_2$ and $z_1$ both lie in $B_{\epsilon_{y_1}} (y_1, D)$, and since $f$ is injective on this ball we have $z_1 = z_2$.

This argument is useful for the construction of equivariant tubular neighborhoods in certain infinite-dimensional settings. See http://arxiv.org/abs/1006.0063; I just updated it so I guess the new version will show up tomorrow. The equivariant version of the above argument is in Proposition 2.3 or 2.4, depending on the version.

`\mapsto`

s in weird places! :) – Mariano Suárez-Álvarez Feb 8 '11 at 20:05Characteristic Classes(and probably most other texts on differential geometry topology). – Dave Anderson Feb 9 '11 at 19:38