# 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 $\N^\circ$, $f(\M)=\N$.

• It should never be an immersion with completely transverse self-intersections. Intuitively for this to happen, you would need to cross the discriminant locus in the space of immersions. But there's lots of easy examples (even for immersed curves in the plane) where the intersection is not transverse. – PVAL-inactive Oct 12 '17 at 18:13
• You assumptions imply that $f_n(M)=N$ for every $n$, i.e. each $f_n$ is a homeomorphism (use the invariance of domain theorem to show that each $f_n$ is an open map). It is a pleasant exercise to show that indeed their limit is again a homeomorphism. – Moishe Kohan Oct 14 '17 at 4:03

## 1 Answer

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.

• @AsafShachar: You can always choose a Riemannian metric on the target space to make the boundary totally geodesic. Then use the "straight-line homotopy" between the maps (provided that $d(f(x), f_k(x))$ is less than the injectivity radius at $x$, use homotopy along the geodesic between these points). This is a very standard trick. – Moishe Kohan Oct 26 '17 at 21:29
• Thanks. I am not sure I understand what happens when $f(x) \in \partial M, f_k(x) \notin \partial M$ or vice versa. (Do you have a unique geodesic between a boundary point and a nearby interior point? I am not sure how making the boundary totally geodesic helps). – Asaf Shachar Oct 29 '17 at 17:05
• @AsafShachar: Yes, the geodesic is unique and it depends continuously on the end-points. (If you know the proof for manifolds without boundary, just repeat it in this setting.) It is simpler if you assume additionally that the metric near the boundary is a product metric $\partial N \times [0,a]$. Do you still have trouble proving uniqueness in this case? – Moishe Kohan Oct 29 '17 at 17:10
• I am sorry to bother you again, but something confuses me: Why are the $f_k$ homotopic to $f$ relative to the boundary $\partial M$? (We do not assume $f_k$ all agree on $\partial M$ so this does not make sense to me). – Asaf Shachar Nov 21 '17 at 16:42
• @AsafShachar: Relative is understood here in a weak sense, namely homotopic as maps of pairs $(M,\partial M)\to (N,\partial N)$. – Moishe Kohan Nov 21 '17 at 22:34