In A Short Exposition of the Madsen-Weiss Theorem, Hatcher discusses (starting at p.6) a basis for a topology on the space $\mathcal{C}^n$, the space of all smooth oriented $d$-dimensional submanifolds of $\mathbb{R}^n$ that are properly embedded. The topology is defined to have as basis the sets $C(B,V)$, where
- B is a closed ball in $\mathbb{R}^n$ centered at the origin.
- V is an open set in the standard $C^\infty$-topology on the space of $d$-dimensional smooth compact submanifolds of $B$ that are properly embedded. Let's call this space $\text{Emb}_c^p(B)$.
- Define $C(B,V) := \{ M \in \mathcal{C}^n \mid \text{$M$ intersects $\partial B$ transversely, } M \cap B \in V \}$.
Then Hatcher states that it is easy to show that, as $B$ and $V$ vary, these sets really define a basis, i.e. we have to show the following:
- $\{ C(B,V) \}$ covers $\mathcal{C}^n$.
- For every $C(B,V)$ and $C(B',V')$ and every manifold $M \in C(B,V) \cap C(B',V')$ there exist $B''$ and $V''$ such that $M \in C(B'',V'') \subseteq C(B,V) \cap C(B',V')$.
I believe I can show (1.), however I am having trouble showing (2.).
In order to show (2.), we simply need to choose suitable $B''$ and $V''$ such that the above holds. Wlog we can assume that $B \subseteq B'$, since all balls were centered at the origin. So my guess is that we choose $B'' := B$. However, I am unsure how to choose $V''$. If my understanding is correct, we cannot simply take $V'' := V \cap V'$, since $V$ is an open set in $\text{Emb}_c^p(B)$ and $V'$ an open set in $\text{Emb}_c^p(B')$, and I believe we can in general write $$ \text{Emb}_c^p(B) = \bigsqcup_{S \in I_B} C^\infty(S,B) $$ where $I_B$ is defined as the set $$\{ S \subseteq B \mid \text{$S$ smooth compact submfld., $S \cap \partial B = \partial S$, intersection is transverse} \},$$ so generally the intersection $V \cap V'$ is not well-defined.
However, I believe I can sort of fix this issue by viewing $V$ as sitting inside $\text{Emb}_c^p(B')$. If this works, it would only remain to show that $M \cap B' \in V' \implies M \cap B \in V'$, so long as $B \subseteq B'$. But again I seem to run aground with this approach. If need be, I can elaborate more on my work up to this point.
Perhaps there is a much more elementary proof, or I am simply misinterpreting some notions. I am grateful for any comments or hints!
