How does a smooth structure on a subset of a manifold determine its status as an immersed submanifold?

As titles are limited to 150 characters, allow me to rephrase my question in a way that is hopefully more precise:

Given a $d$-dimensional smooth manifold $M$ and some $k$-dimensional subset $S\subset M$, how does a smooth structure on $S$ turn $S$ into an immersed submanifold (or alternatively, prevent $S$ from being an immersed submanifold)?

I have the following definition: $S\subset M$ is an immersed submanifold if there exists a smooth manifold structure on $S$ such that the inclusion $i:S\hookrightarrow M$ is a smooth immersion. Continuing, $i$ is a smooth immersion if $i$ is a smooth map between $S$ and $M$ (which in this case, it must be) and $di_a: T_aU\to T_{i(a) }M$ is injective for all $a\in U$.

At first glance, I have a difficult time seeing where the smooth structure you build on $S$ even comes into play (just to be more clear about what I mean when I just say "building a smooth structure on $S$", I mean to give $S$ with some atlas of smooth charts $\varphi_i:S\to\mathbb{R}^k$). Though after giving it some thought, the smooth structure you build determines a topology on $S$, and as $i$ is continuous, you would need the inverse images of open sets in $M$ to be open in $S$, something which would need to be facilitated by the smooth structure you pick. Is this the correspondence between the two notions, or is there something more I'm completely missing?

In summary, my thoughts on the matter are as follows:

The smooth structure you build determines a topology on the subset $S$, and you need this topology to satisfy certain properties in order to allow $i$ to be a smooth immersion.

Any input is welcome!

For simplicity, let's say that the ambient manifold $M = \mathbb{R}^k$, Euclidean space. We have a subset $S \subset \mathbb{R}^k$, and we want to investigate meaningful ways of endowing $S$ with a smooth submanifold structure. The word "submanifold" here suggests, as you rightly point out, that something about the smooth structure on $S$ needs to be compatible with the smooth structure on $\mathbb{R}^k$. However, the compatibility is not simply topological, since there are some conditions on the derivative of the inclusion $i$.
To give an example, let $S$ be a square (not filled in) sitting inside $\mathbb{R}^2$. It's not hard to give the square a smooth structure as a abstract manifold, since nothing about the abstract, topological description of a square suggests its sharp, pointy corners. You can even do this in a way where the natural inclusion $i: S \to M$ is continuous. However, it is not possible to have $di$ be injective at a corner. Using the regular parametrization of the square, the derivatives fail to exist at the corner. If we were a little more clever, we could have our parametrization slow down asymptotically near the corners, so that the derivatives agree as we come along either side -- still, this will force the derivative to be zero there, so $di_{corner} =0$ is not injective.
The above example should shed some light on the general situation. The derivative of the inclusion $i: S \hookrightarrow M$ depends on the parametrization we put on $S$, since the derivative of a smooth map between smooth manifolds is defined in terms of our charts. If we put a smooth structure on $S$ that entirely ignores the way $S$ sits inside $M$ (not just topologically, but also in terms of "differential" phenomenon like cusps), there is no knowing what might happen when we transfer everything back to Euclidean space via our charts and take the derivative. It might fail to exist or not be injective. The condition that $di$ be injective forces us to choose a smooth structure on $S$ that at least contains the restriction of the charts on $M$ to $S$, and moreover that those restrictions are smooth, without corners or singular differential.
• Good answer. I would add the term local representation, though, since this is the point where the structure on $S$ is explicitly relevant. – gofvonx Nov 15 '13 at 9:05