Is the dimension of an immersed manifold unique?

Question

Let $F:M^n \rightarrow \mathbb{R}^m$ be an immersion between two manifolds which is not necessarily injective. Denote the immersed manifold by $S = F(M^n)$, we are tempted to say that $S$ has dimension $n$. However is it possible that there exists a second immersion $G:\tilde{M}^{n+1} \rightarrow \mathbb{R}^m$ such that $G(\tilde{M}^{n+1}) \subset S$? We are assuming that all manifolds are second countable.

Answer 1

Let $K^k, M^m, N^n$ be topological manifolds of dimensions $k, m, n$ respectively, such that $k< m$. (All my manifolds are 2nd countable.) Actually, it suffices to assume that $N$ is a Hausdorff topological space, I will not need it to be a manifold.

Suppose that $$ f: K\to N, g: M\to N $$ are locally injective continuous maps.

Proposition. The image of $f$ cannot contain the image of $g$.

Proof. Since $f$ is locally injective and $K$ is 2nd countable, there exists a countable cover $\{U_i: i\in I\}$ of $K$ such that $f|_{cl(U_i)}$ is 1-1 (hence, a homeomorphism to its image) for each $i$. Similarly, we have a countable cover $\{V_j: j\in J\}$ of $M$ such that $g|_{cl(V_j)}$ is 1-1 for each $j$. For each $i\in I, j\in J$ consider the compact subset $$ L_{ij}= cl(V_j)\cap g^{-1}( f(cl(U_i))\subset M. $$ This subset is homeomorphic to a compact subset of $K$ (via a composition of the restriction of $g$ to $cl(V_j)$ followed by the inverse to $f|_{cl(U_i)}$). Hence, since $dim(K)< dim(M)$, by the invariance of domain theorem $L_{ij}$ has empty interior in $M$. (Here the proof is easier in the smooth setting since $L_{ij}\subset M$ is then diffeomorphic to a compact subset of $K$.) It follows that $$ X:=\bigcup_{i\in I, j\in J} L_{ij} $$ has empty interior in $M$ by the Baire Category Theorem. In particular, $g^{-1}(f(K))\ne M$, hence, $g(M)$ is not contained in $f(K)$. qed

Answer 2

Let $F:M^n\rightarrow \mathbb{R}^m$ be an immersion and assume that $m>n$. Suppose that there is some point $y\in\mathbb{R}^m$ such that $$ \bigcup_{x\in F^{-1}(y)}dF_x(T_xM) $$ is a subspace of $T_yN$ of dimension (strictly) greater than $n$. This implies that $x\in F^{-1}(y)$ has an uncountable number of points.

If $F^{-1}(y)$ is discrete, this contradicts second-countability. If $F^{-1}(y)$ is not discrete, then it contains a limit point $x_0$. But then $F$ is not an immersion since, at $x_0$, $F$ is not locally injective. This is a contradiction.

If $G$ were an immersion whose image is included in the image of $F$, then suppose that $G(z)=y$ for some $y\in F(M)$. Since $G$ is an immersion, $dG_z$ is injective and the tangent space is $n+1$-dimensional. This means that you have $n+1$ independent "directions" that you can travel. On the other hand, the union above includes at most countably many $n$-dimensional directions. This is fewer than for $G(z)$, which shows that you can travel in a direction that remains in the image of $G$, but not in the image of $F$.

You can make this discussion of directions rigorous by identifying tangent vectors with curves through a point in your manifold.