Topology of an immersed submanifold

Question

Problem somewhat related to this post, and I refer to the same exercise in John Lee's Introduction to Smooth Manifolds.

My issue is trying to understand the relationship between a general subspace topology, and the topology given to an immersed submanifold.

Let $ M$ be a smooth manifold and $S \subseteq M$ be an immersed submanifold of $M.$ Following the definition outlined in Lee, it is equipped with a topology such that $S$ is a topological smooth manifold and the inclusion map $ \iota :S \hookrightarrow M$ is a smooth immersion.

The problem: We want to show that open sets in $S$ with respect to the subspace topology are also open with respect to the given submanifold topology.

My understanding is as follows: For the subspace topology, a set $U \subseteq S$ is open if it can be expressed as $U = S \cap V$ for some open set $V \in M$. Following guidance from the linked post, since the inclusion map $\iota$ is smooth, it is continuous, and hence the preimage preserves open sets. If $V$ is open in $M$ (with respect to $M$'s topology), then $\iota^{-1}(V) = V \cap S$ is open; but here is where I am confused.

My questions: With respect to what topology would $V \cap S$ would it be open in?

Is showing that $U =V \cap S$ is open as preimage of an open set under the inclusion map what is meant by being "open" with respect to the submanifold topology (in particular, for an immersed submanifold)?

If not, what mechanism do we use to prove that a set is open in this topology (for instance, in the subspace topology, we must show there exists an open $V$ in $M$ such that $U = S\cap V.$ How does one do this in the submanifold topology, in order to prove the statement?)

Like how the open sets of the subspace topology can be found by intersecting $S$ with every open set in $V$, that is

$\mathcal{O} = \{U \subseteq S : \text{exists an open set } V \subseteq M \text{ such that } U = V \cap S\}$,

can one write something similar for the submanifold topology?

I would appreciate any clarification on this!

Answer 1

$S$ is a subset of $M$. Then the set $S$ is endowed with a topology $\tau_S$ and smooth structure, a priori independent from $M$. This topology $\tau_S$ is the submanifold topology.

As $S$ is a subset of $M$ there is the inclusion map $\iota: S \to M$. By definition of immersed submanifold $\iota$ is now a smooth immersion and in particular continuous. Thus $\iota^{-1}(\tau_M) \subseteq \tau_S$, where $\tau_M$ is the topology of $M$.

But the subspace topology $\tau_{S\subseteq M}$ is, by definition, $\iota^{-1}(\tau_M) = \tau_{S\subseteq M}$. Hence $\tau_{S\subseteq M} \subseteq \tau_S$.

Answer 2

• First remark: let $X$ be a set, $(Y,\tau_Y)$ a topological space and $f:X\to Y$ any function. Then, the collection $f^*(\tau_Y):= \{f^{-1}[V]\,:\, V\in \tau_Y\}$ is a topology on $X$. This is an easy enough exercise for you to verify. In particular, if $X$ is a subset of $Y$ and $f=\iota:X\hookrightarrow Y$ is the canonical inclusion map, then $\iota^*(\tau_Y)$ is called the subset topology on $X$ inherited from $Y$.

• Second remark: Let $(X,\tau_X), (Y,\tau_Y)$ be topological spaces and $f:X\to Y$ a given function. Then, by definition, we will have that $f$ is continuous if and only if $f^*(\tau_Y)\subseteq \tau_X$. But note that the inclusion could very well be strict. We say that $f$ is a topological embedding if $f$ is injective and $f^*(\tau_Y)=\tau_X$ (this is equivalent to saying the target-restricted map $f:(X,\tau_X)\to (f[X], \tau_{\text{subset},\,\, f[X]\subset Y})$ is a homeomorphism, or equivalently, that it has a continuous inverse).

Now, let $M$ be a smooth manifold and $S\subseteq M$ an immersed submanifold. By definition this sentence means we have $(M,\mathscr{A}_M,\tau_M)$, a set a maximal smooth atlas and a topology on $M$, and likewise $(S,\mathscr{A}_S,\tau_S)$ for $S$, such that the inclusion map $\iota:S\hookrightarrow M$ is a smooth immersion. Note that the choices of $\mathscr{A}_S,\tau_S$ are not uniquely determined by $(M,\mathscr{A}_M,\tau_M)$, meaning that you have to specify this for us before we can proceed with the discussion; see here.

Since $\iota$ is smooth, it is in particular continuous, so by our previous two remarks, we have \begin{align} \tau_{\text{subset},\, S\subseteq M}:=\iota^*(\tau_M)\subseteq \tau_{S}. \end{align} So, the fact that every set which is open in the subset topology is also open relative to the given immersed submanifold topology is pretty much trivially true by definition.

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At this stage, I should remark that the concept of an immersed submanifold is unnecessarily confusing. I find it much cleaner to just speak of immersions/ injective immersions.

For example, the preceding discussion could have been phrased as follows: suppose $(X,\mathscr{A}_X,\tau_X), (Y,\mathscr{A}_Y,\tau_Y)$ are smooth manifolds and $f:X\to Y$ is a smooth injective immersion. Then, we have \begin{align} f^*(\tau_Y)\subseteq \tau_X, \end{align} but the inclusion may be strict (if they happen to be equal, then $f$ is said to be a smoth embedding).

Answer 3

Im a beginner myself so i could be wrong, $\iota$ is a smooth immersion so $S$ is already equipped with the submanifold topology as the submanifold topology is just the topology that makes $\iota$ into a smooth immersion. Because $\iota$ is continuous, if a set $V\cap S$ is open in the subspace topology, its preimage under $\iota$ is open in the submanifold topology.