I know I must be missing something somewhere; at the risk of being laughed at (a lot) I am asking if someone could point out to me exactly where. The question has to do with submanifolds, topology and open maps.
Say I have an open subset of $R^n$, call it $P$, topology is the relative one. Further, I have an $(n-1)$-dimensional embedded submanifold $N$ of $P$, so its topology is the relative one as well. Now suppose there is a smooth map $\lambda : P \rightarrow R^{n-1}$ whose restriction to $N$ is injective and is of rank $n-1$ everywhere. Consequently, as I understand it, $\lambda$ is an open map, taking every open (in the topology of $N$) set to an open (in the topology of $R^{n-1}$) set. This should also (by complementation) mean that $\lambda$ takes closed sets to closed sets. So here's my question: since $N$ is both open and closed in its own (the relative) topology, shouldn't the image $\lambda(N)$ be both open and closed in $R^{n-1}$? For separate reasons I know this result cannot be true without further conditions, but I can't see my mistake. Is it that I'm not being precise enough about the topology on $R^{n-1}$; e.g. it's a tautology that $\lambda(N)$ is both open and closed in $\lambda(N)$? But then what does an "open map" mean? Only that $\lambda(V)$ is open in $\lambda(N)$ whenever $V$ is open in $N$? That really doesn't help if we want to infer that $\lambda(N)$ is open in $R^{n-1}$.
Sorry for the rambling but my head hurts.
Edit: It appears this question is intimately connected to the concepts of relatively open map and strongly open map. In particular, I'm not certain which one of these $\lambda$ turns out to be. I will clearly need to do more research, although there doesn't seem to be much available on these particular topics.
An open map $f:X\to Y$ between spaces $X$ and $Y$ is not necessarily closed. Recall that, in general, $f(X\setminus A)\neq f(X)\setminus f(A)$, even if $f$ is injective (injectivity only implies $f(X\setminus A)\subset f(X)\setminus f(A)$). In any case, complementation won’t take you anywhere.
And what I said is true: there are a lot of examples of open (even injective) maps which are not closed. A simple one, which is similar to the situation you are facing, is the inclusion map from $(0,1)$ into $\mathbb{R}$: $$j:(0,1)\to\mathbb R,\ j(x)=x$$ It is open, since any open subset $U$ of $(0,1)$ is open in $\mathbb{R}$, and $j(U)=U$, but it cannot be closed, since $(0,1)$ is closed in $(0,1)$ and $j(0,1)=(0,1)$ is not closed in $\mathbb{R}$.