@BalarkaSen @Thorgott On second reading, I've realised that I don't know why one knows that $M_g$ is generated by the $n$ elements $\{m_i\}_{i=1}^n$ as an $A_g$-module. I can see that the generators of $M$ map to elements in $M_\mathfrak{p}$ that can be expressed in the form $\sum a_i\cdot m_i/ g$ for $a_i\in A$, but even still, we end up allowing $A_\mathfrak{p}$-linear combinations of these generators (in $M_\mathfrak{p}$)
(I.e. it's not clear to me that one can use the first half of their argument to show that $M_\mathfrak{p}$ treated as an $A_g$-module by restriction is isomorphic to $M_g$, which I think they are using?)
@Astyx Sorry can you be more precise by saying $A_\mathfrak{p}$-linearly generated, or whichever ring you mean (and/or say "every element of <module>")
@Astyx Note that there are k of these, and we want to reduce this to n in the potentially-exact-sequence in question, and we only re express these in terms of the basis of $M_\mathfrak{p}$ as an $A_\mathfrak{p}$-module, which is part of my contention
because every element in $M_g$ is of the form $\sum {a_i\over g^n}m_i$, because the $b_j$ generate M as an A module, hence generate $M_g$ as an $A_g$ module
I agree that that computation makes sense in terms of equivalence classes occurring in defining $M_\mathfrak{p}$ as an $A_\mathfrak{p}$-module, and I would also agree that this makes sense in $M_g$ if I knew that extending and then restricting yields an isomorphism
Like formally their computation occurs inside $M_\mathfrak{p}$
Like that equality apriori means equivalence in $M_\mathfrak{p}$
Or in other words I agree that this solves the problem if the unit of adjunction for $f_!\leftrightarrow f^*$ is a natural isomorphism
@tigre pick an $A_\mathfrak p$-basis of $M_\mathfrak p$ that lives in $M$. This gives you an map $A^n \to M$ that is an isomorphism at $\mathfrak p$. Take the kernel and cokernel to be $K$ and $Q$ and form an exact sequence $0 \to K \to A^n \to M \to Q \to 0$ that becomes $0 \to 0 \to A_\mathfrak p^n \to M_\mathfrak p \to 0 \to 0$, so $K_\mathfrak p = 0$ and $Q_\mathfrak p = 0$, so there is $g_1$ and $g_2$ such that $K_{g_1} = 0$ and $Q_{g_2} = 0$. Now take $f = g_1 g_2$.
$\newcommand{\gp}{{\mathfrak p}}$Let $\left\{ \frac{m_1}{s_1}, \cdots, \frac{m_n}{s_n} \right\}$ be an $A_\gp$-basis of $M_\gp$, where $m_i \in M$ and $s_i \in A-\gp$. Observe that $\{m_1, \cdots, m_n\}$ is still a $A_\gp$-basis of $M_\gp$.
This gives an map $A^n \to M$ that localizes to an isomo...
@user83244835 If people ignore you, they probably don't have a good answer (or the effort to answer is too high for the quality of the proposed answer, etc)
Well, yes, in general, you are asking for the largest possible domain the function can have (after you've analytically continued as much as you can). For example, if you have a power series with radius of convergence $1$, then you know that there is at least one point on the circle of radius $1$ where you cannot analytically continue. $f(z)=1/(1-z)$ is an example, but it's analytic everywhere else.
But there are famous power series with radius of convergence $1$ that cannot be analytically continued across the circle at all.
Doing all this geometry has been fun and it is definitely adding a dimension to how I look at things......and as you'll love to hear......creating a picture of why the algebra makes sense.
@robjohn: Stop your silliness. Grr. Someone just downvoted for no apparent reason.
@user2103480: Here's an example. Suppose you have an embedded submanifold $M\subset X$. You have a smooth function on $M$. You wish to extend it to be a smooth function on all of $X$.
Another basic application is that a tubular neighborhood is diffeomorphic to a neighborhood of the zero section of the normal bundle of the submanifold. So you can transfer, for example, intersection theory questions in the normal bundle to intersection theory questions in the ambient manifold.
in the same spirit, tubular neighborhood thm tells you that the homology of the normal bundle relative complement of zero section is isomorphic to the homology X relative complement of M
@TedShifrin Nothing looks bad there. I think someone is downvoting for other reasons. Not the way we ask people to vote, but it's hard to control since there is no requirement to comment when downvoting. I have a lot of unwarranted downvotes.
I should say seemingly unwarranted; it's hard to tell for sure without a comment.
The only thing that I can see that someone might disagree with is the statement: "You should always prefer to do a volume integral rather than a surface integral."
Other than that, there is nothing that seems opinion based.
@robjohn: I almost put "almost" in that sentence. But I added the phrase about satisfying hypotheses instead. It's sound pedagogical advice, not opinion :P
Hm. One thing we used was substitution of the tubular neighborhood into complements. Is the trick here that M\N is not necessarily a manifold for N embedded into M, but M\VN is?
I have a continuous surjection $f:X\to Y$ between compact spaces. For $B\subseteq X$ open let the fiber image $f_{fib}(B)$ be $\{y\in Y\mid f^{-1}(y)\subseteq B\}$. Apparently this fiber image is always open (possibly empty)
I have an argument thinking about convergence of the fibers in the hyperspaces of closed sets of $X$, but that's nonsense, it should be completely elementary
@robjohn @TedShifrin can it not be integrated into the system that a reason (which involves a character minimum) be implemented in terms of down voting?...I guess it would have to go hand in hand with individuals asking for help to explicitly show the work they've done and have the attempt have substance...
As I said, the tube lemma shows up numerous times in algebraic and differential topology, and things like dynamical systems, so I consider it fundamental.
Even the algebraic geometers know it, under the name "universally closed"
@AlessandroCodenotti Yes this is straightforward tube lemma which can equivalently be stated as: for all Y and compact X the projection X x Y -> Y is a closed map
Your set f_fib is best characterized by its complement: f_fib(B)^c = {y | exists x in B^c with f(x) = y} = p_2({(x, f(x)) | x in B^c}). Notice that {x, f(x) | x in B^c} is a closed subset of X x Y: the graph of f is closed because Y is Hausdorff and f is continuous, and this is its intersection with the basic closed set B^c x Y
Since p_2 is a closed map f_fib(B)^c is a closed set hence f_fib(B) is open
This is an abstract phrasing of a more concrete argument where you show f_fib(B) is locally open