Restrictions of a local diffeomorphism I am wondering if a local diifeomorphism has the following property (prove or disprove):
Let $M,N$ be differentiable manifolds, and $f:M \to N$ be a local diffeomorphism. Suppose $Z$ is a closed subset in $N$ such that the restriction
$$
f|_{f^{-1}(Z)}: f^{-1}(Z) \to Z
$$
is a bijection. Then there exists an open neighborhood $U$ of $f^{-1}(Z)$ in $M$ such that the restriction $$f|_U: U \to f(U)$$ is a diffeomorphism.
Thank you!
 A: Truly speaking I do not understand the 4 downvotes in order to close this question.
It is not that trivial (at least I tried 4 approaches till found a satisfactory solution).
This question has an affirmative answer if the manifolds $M$ and $N$ are paracompact (usually this requirement is included to the definition of a manifold).
We shall use the following known fact (see Corollary 8.1.11 in "General Topology" of Engelking).

Lemma. For any open cover $\mathcal U$ of a paracompact space $X$ there exists a continuous pseudometric $d$ on $X$ such that for every $x\in X$ the unit ball $B_d(x;1)=\{y\in X:d(x,y)<1\}$ is contained in some set $U\in\mathcal U$.

Fix a local diffeomorphism $f:M\to N$ between manifolds and a closed set $F\subset N$ such that $f{\restriction}f^{-1}(F):f^{-1}(F)\to F$ is bijective.
By the Lemma, the space $M$ admits a continuous pseudometric $d$ such that for every $x\in X$ the map $f{\restriction}B_d(x;1):B_d(x;1)\to f(B_d(x;1))$ is a diffeomorphism onto an open subset $f(B_d(x;1))$ of the manifold $N$. Applying the Lemma to the open cover $\{f(B_d(x;\frac12)):x\in M\}$ of $N$, find a continuous pseudometric $\rho$ on $N$ such that each unit ball $B_\rho(y;1)$, $y\in N$, is contained in the image $f(B_d(x;\frac12))$ for some $x\in X$. Here $B_d(x,\frac12)=\{y\in X:d(x,y)<\frac12\}$ is the ball of radius $\frac12$.
Now consider the open neighborhood $$V=\{v\in X:\exists x\in f^{-1}(F),\;\max\{d(x,v),\rho(f(x),f(v))\}<\tfrac12\}$$ of $f^{-1}(F)$.
We claim that $f{\restriction}V$ is injective and hence a diffeomorphism.
Given any points $u,v\in V$ with $f(u)=f(v)$, we should prove that $u=v$. For the points $u,v$ choose points $x,y\in f^{-1}(F)$ such that $\max\{d(x,u),d(y,v),\rho(f(x),f(u)),\rho(f(y),f(v))\}<\frac12$. By the choice of the pseudometric $\rho$, the unit ball $B_\rho(f(u))$ is contained in $f(B_d(z;\frac12))$ for some $z\in M$. Taking into account that $f(u)=f(v)$ and $\max\{\rho(f(x),f(u)),\rho(f(y),f(v))\}<\frac12<1$, we conclude that $f(x),f(y)\in B_\rho(f(u);1)\subset f(B_d(z;\frac12))$. Since $f{\restriction}f^{-1}(F)$ is injective, and $f{\restriction}B_d(z;1):B_d(z;1)\to f(B_d(z;1)$ is bijective, $x,y\in f^{-1}(F)\cap B_d(z;\frac12)$. Taking into account that $\max\{d(x,u),d(y,v)\}<\frac12$, we conclude that $u,v\in B_d(z;1)$. Now the injectivity of the restriction $f{\restriction}B_d(z;1)$ yields the desired equality $u=v$.
