Is uniform limit of embeddings injective? $\newcommand{\pl}{\partial}$
$\newcommand{\M}{\mathcal{M}}$
$\newcommand{\N}{\mathcal{N}}$
Let $\M,\N$ be smooth compact, connected, oriented manifolds of the same dimension with non-empty boundaries.
Let $\,f_n:\M \to \N$ be a sequence of smooth orientation-preserving embeddings, satisfying $f_n(\partial \M) \subseteq \partial \N$. Suppose that $f_n$ converges uniformly to a smooth immersion $f:\M \to \N$. 

Is it true that $f$ is injective?

Edit: 
It turns out the assumptions above imply that $f_n$ are diffeomorphisms, and hence $f$ is surjective. (A proof is provided below, if you are interested). Thus, the question becomes:

Suppose $f_n:\M \to \N$ are diffeomorphisms between manifolds with boundary, which converge uniformly to a smooth surjective immersion $f$. Is $f$ injective?


A proof the $f_n$ are diffeomorphisms:
Lemma: Let $f:\M \to \N$ be a smooth immersion, and suppose that $f(\pl \M)\subset \pl \N$ Then $f$ is surjective.
Applying this lemma to the $f_n$, we get that they are surjective embeddings, hence diffeomorphisms.
A proof the lemma:
Since $f:\M \to \N$ is smooth and $df$ is invertible, any interior point $x\in \M^\circ$ is mapped to an interior point of $\N$, hence $f(\M^\circ) \subset \N^\circ$. 
By the inverse function theorem $f:\M^\circ \to \N^\circ$ is a local diffeomorphism and in particular  an open map, so $f(\M^\circ)$ is open in $\N^\circ$. We proceed to show it's also closed (in $\N^\circ$).
Now, let $y_n=f(x_n)\in f(\M^\circ)$ converges to $y\in \N^\circ$. 
Since $\M$ is compact and $f$ is continuous, we may assume, by taking a subsequence,  that $x_n\to x\in \M$, and $y=f(x)$.
Since $f(\pl\M) \subset\pl\N$ and $y\in\N^\circ$, it follows that $x\in \M^\circ$, i.e. $y\in f(\M^\circ)$, which implies that $f(\M^\circ)$ is closed in $\N^\circ$.
Thus, we showed that $f(\M^\circ)$ is clopen in $\N^\circ$.
Since $\N^\circ$ is connected, $f(\M^\circ)=\N^\circ$. Since $f(\M)$ is closed in $\N$ ($\M$ is compact), and contains the dense subset $\N^\circ$, $f(\M)=\N$.
 A: First of all, the limiting map will satisfy the property that $f(\partial M)\subset \partial N$. Therefore, it will induce a homomorphism 
$$
f_*: H_m(M,\partial M)\to H_m(N, \partial N),
$$
where $m$ is the common dimension of $M$ and $N$. Recall that the degree $deg(f)$ of $f$ is the number $d$ such that 
$$
f_*([M,\partial M])= d [N,\partial N]
$$
where the bracket denotes the relative fundamental class (the generator of the top relative homology group of a connected oriented manifold). Note also that the degree is preserved by relative homotopy and also that the maps $f_k$, $f$ are homotopic rel. boundary of $M$ for large $k$. Since each $f_k$ is a diffeomorphism, its degree equals $1$, hence, $f$ also has degree 1. Recall also that the degree of a smooth map $f$ can be computed as
$$
\sum_{x\in f^{-1}(y)} (sign(det(df_x)),
$$ 
where $y\in N$ is a regular value of $f$ (in your case, an arbitrary point since $f$ is assumed to be an immersion). Moreover, since $f$ is an immersion and $M$ is connected, $sign(det(df_x)$ is independent of $x$, say, equals $+1$. Hence, if $f$ is not injective, then $deg(f)>1$, which is a contradiction. For the same reason, $f$ is surjective since otherwise $deg(f)=0$ as we can take $y\notin f(M)$. 
You can find all the background material for instance in "Differential Topology" by Guillemin and Pollack. 
