Is the image of a continuous functions always a topological manifold

Question

Let $f:\mathbb{R}^n\rightarrow \mathbb{R}^m$ be continuous. Then, for some sufficiently large $M\geq m$, can we always guarantee that $f(\mathbb{R}^n)$ is locally-homeomorphic to some $\mathbb{R}^N$, where $N\leq n$?

My intuition is no, since I can imagine a curve crossing itself; but I can't write down an explicit example.

Answer 1

No. Take, for instance,$$\begin{array}{rccc}f\colon&\Bbb R&\longrightarrow&\Bbb R^2\\&t&\mapsto&\left(\frac{\cos(t)}{1+\sin^2(t)},\frac{\sin(t)\cos(t)}{1+\sin^2(t)}\right).\end{array}$$Its range is a Lemniscate of Bernoulli, which is not locally homeomorphic to $\Bbb R$ (precisely because of the existence of a self-intersection).

Answer 2

Another counterexample with a different issue: Let $$f:\mathbb R^2\to\mathbb R^2,(x,y)\mapsto\begin{cases}(x,y)&x<0\\ (x,(1-x)y)&x\in[0,1]\\(x,0)&x>1\end{cases}$$ For $x<1$, the image is locally homeomorphic to $\mathbb R^2$. For $x>1$, it is locally homeomorphic to $\mathbb R$. At the boundary of both regions, $x=1$, the image can't be locally homeomorphic to either.