Proposition 1

Let $\varphi : X\rightarrow Y$ be a morphism of class $C^r$, A a compact subset of X. The following conditions are equivalent:

(i) The restriction of $\varphi$ to A is injective, and $\varphi$ is an immersion at every point of A;

(ii) there exists an open neighbourhood U of A such that $\varphi$ induces an embedding of U into Y.

When these conditions are satisfied, $\varphi$ is said to be an embedding in the neighbourhood of A.

We prove that (i) implies (ii), the other implication being clear. Assuming (i), there exists an open neighbourhood V of X containing A such that the restriction of $\varphi$ to V is an immersion (Differentiable and Analytic Manifolds, Results, 5.7.1). Denote by $\Gamma$ the set of points $(x, y)$ in $V\times V$ such that $\varphi (x) =\varphi (y)$, and by $\Delta$ the diagonal in $V\times V$. Then $\Delta$ is an open subset of $\Gamma :$ indeed, for all $x\in V$, there exists an open neighbourhood $U_x$ of $x$ such that the restriction of $\varphi$ to $U_x$ is injective, that is, such that $\Gamma \cap (U_x\times U_x) =$ $\Delta \cap (U_x\times U_x)$.

Since Y is separated, $\Gamma$ is closed in $V\times V$; consequently the complement W of $\Gamma$ -- $\Delta$ in $V\times V$ is open. By assumption, W contains $A\times A$; Lemma 1 implies that there exists an open subset $U'$ of V containing A such that $U'\times U'\subset W$, that is, such that the restriction of $\varphi$ to $U'$ is injective. Moreover, there exists an open neighbourhood U of A whose closure is compact and contained in $U'($General Topology, Chap. I, §9, no. 7, Prop. 10). Then $\varphi$ induces a homeomorphism from U to $\varphi$(U), and consequently from U to $\varphi (U)$; it follows that the restriction of $\varphi$ to U is an embedding (Differentiable and Analytic Manifolds, Results, 5.8.3).