First of all, I'm still not completely familiar and comfortable with all the vocabulary, so it could be that my question doesn't make complete sense.
I'm reading through Loring Tu's "An Introduction to Manifolds", and while I haven't completely understood everything I've read so far, I got to one paragraph that has me completely stuck. Below is the paragraph in question:
"Suppose $f: N\to M$ is a $C^{\infty}$ map whose image $f(N)$ lies in a subset $S\subset M$. If $S$ is a manifold, is the induced map $\tilde{f}\to S$ also $C^{\infty}$? This question is more subtle than it looks, because the answer depends on whether $S$ is a regular submanifold or an immersed submanifold of $M$."
This might be a rather oversimplified viewpoint, but imaging $f$ as a function from $\mathbb{R}^m\to\mathbb{R}^n$, the induced map is simply a restriction of the codomain of $f$. So why should the smoothness of the map $\tilde{f}: \mathbb{R}^m\to S\subset\mathbb{R}^n$ depend on the relative structure of $S$ and the parent space (in my example $\mathbb{R}^n$)?
Or another way I look at it; since $S$ is defined to be the image $f(N)$, that means that for every point $n\in N$, $f(n)\in S$. This fact, combined with the fact that $f$ is $C^{\infty}$, seems like a sufficient condition for $\tilde{f}$ to be smooth. What is wrong with this line of reasoning?
But that's not the definition of what it means for $\tilde{f}$ to be smooth. By definition, $\tilde{f}:N\to S$ is smooth if it gives a smooth map between every pair of charts on $N$ and $S$. Since $f$ is smooth, we know that $f$ gives a smooth map between every pair of charts on $N$ and $M$. But charts on $M$ and charts on $S$ are not the same thing! Moreover, the charts on $S$ depend not only on $S$ as a subset of $M$, but on the specific smooth manifold structure we choose to put on $S$. (In this way it is extremely misleading to say something like "If $S$ is a manifold"; it's not that $S$ is a manifold, but that we choose some manifold structure on $S$, which is not unique.)
In particular, if we pick some totally random smooth structure on $S$ that is unrelated to $M$, there is absolutely no reason to think $f$ should be smooth with respect to the charts on $S$. Now, it appears that the quote you are referring to means to consider only smooth structures on $S$ for which the inclusion map $S\to M$ is an immersion. But even that is not enough. For instance, suppose $M=\mathbb{R}^2$ and $S$ is a "figure 8", given the smooth structure which identifies it with an open interval which starts at the center of the 8, goes around the bottom loop counterclockwise, then goes around the top loop clockwise. If $f:\mathbb{R}\to M$ is a map which instead goes around the bottom loop clockwise and then the top loop counterclockwise, then $\tilde{f}:\mathbb{R}\to S$ is not even continuous: when it finishes the bottom loop, it jumps discontinuously between two separate parts of $S$ in our chart that identifies $S$ with an open interval.