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comment Compact operator
@MatthewDaws In fact, it's a verbatim of Conway's A Course in Functional Analysis, Chapter II, section 5, exercise 6.
Dec
13
comment pseudo inverse of a finite-to-one continuous map and measurability
Remark 3 seems wrong. Consider the Cantor set $\mathcal C\subseteq[0,1]$, and a nonmeasurable set $\mathcal N\subseteq[0,1]$ such that there's a bijection $f\colon\mathcal C\to\mathcal N$. We extend $f$ to $[0,1]$ such that $f(x)=x+1$ for all $x\not\in\mathcal C$. Note that the totality of $y$ such that $\#f^{-1}y=1$ isn't measurable. I doubt we should assume that $\pi$ is continuous.
Dec
13
comment Two generating meromorphic functions seperate points on a compact Riemann surface?
@wisefool So what?
Nov
19
comment Background for reading Milnor's Morse Theory book
@tessellation Is it really self-contained? For example, it seems to me that in section 5, something on homology theory is used, and section 7 needs acquaintance with (relative) homotopy groups.
Nov
19
comment $\left\{\,f\in L^1[0,1]\,\big\vert\,\int_0^1\lvert f\rvert^2>1\right\}$ is open
I don't understand your Mazur-theorem argument, but the result follows from the fact that $L^2$ is reflexive and therefore by Banach-Alaoglu theorem, the unit ball is weakly compact, thus a lower semicontinuous function attains its minimum. The later perturbation argument could be simply replaced with applying Cauchy-Schwarz inequality on the set $\{\,x\,\vert\,g(x)<f(x)-\epsilon\,\}$, I think, and the fact that the functional takes minimum is also a consequence of this, without appealing to the general minimum-existence theorem.
Nov
15
comment Is the $\sigma$-finiteness condition necessary to ensure that $L^p(\mu)$ is reflexive?
It should be Folland's book. In addition, $L^1$ isn't reflexive in general even if $\mu$ is $\sigma$-finite, say, $(L^1(\mathbb R))^*=L^\infty(\mathbb R)$.
Oct
29
comment How to show that a measurable function on $R^d$ can be approximated by step functions?
@Groups Edited, thanks!
Oct
25
comment Proper and free action of a discrete group
Thanks a lot! It seems that Lee's book on smooth manifolds also takes this as definition.
Sep
11
comment Is every local ring a valuation ring?
But, $F$ isn't the field of fractions of $\{0,1\}\subseteq F$?
Sep
11
comment Difference between two concepts of homotopy for simplicial maps?
@ZhenLin It seems that you are right, if the definition of the product is essentially same as the one in Goerss & Jardine. It might be a mistranlation from Russian.
Sep
11
comment Difference between two concepts of homotopy for simplicial maps?
@QiaochuYuan Simply homotopy of $f,g\colon X\to Y$ here, informally speaking, is just a simplicial map $F\colon\Delta[1]\times X\to Y$ from a cylinder $\Delta[1]\times X$ (triangulated canonically, where $\Delta[1]$ is just the simplicial set associated with $I=[0,1]$, i.e., $1$-simplex) to $Y$, both of which are simplicial sets. Sorry for my ignorance. I don't know what you're referring to. Thanks, anyway.
Aug
27
comment 6.17 Theorem : Show that $f \ \ \in \mathfrak R(\alpha)$ if and only if $ f\alpha' \ \ \in \mathfrak R$ ( walter rudin)
It follows from Rudin's argument that the upper and lower (Darboux) integrals of $fd\alpha$ and $f\alpha'dx$ are the same.
Aug
27
comment Differential identity and wedge products
$f\omega$ is just $f\wedge\omega$ where $f$ is considered as a $0$-form, and $d(\omega_1\wedge\omega_2)=d\omega_1\wedge\omega_2+(-1)^p\omega_1\wedge d\omega_2$ where $\omega_1$ is a $p$-form.
Aug
27
comment $f\in C(\mathbb{R})$. What does it mean?
But it's usually denoted as $C^n(\mathbb R)$ or $\mathcal C^n(\mathbb R)$ as well as $\mathcal C^0(\mathbb R)$ without parentheses.
Aug
27
comment A question on short exact sequences.
I found learning diagram chasing isn't that easy for a beginner from books, but there're two materials which seems more accessible: Hatcher's Algebraic Topology, pp115, subsection Relative Homology Group, and Eisenbud's Commmutative Algebra, pp637, subsection A3.7.
Aug
26
comment Differentiability of non-analytic complex functions
Well, and by Looman-Menchoff theorem, a continuous complex function is holomorphic in some region if and only if it satisfies Cauchy-Riemann equations everywhere.
Aug
24
comment Homology of a finite graph follows from Mayer-Vietoris sequence?
@LeeMosher Thanks. I've edited to include these.
Aug
20
comment Cover a sphere by two closed subsets not containing a closed self-antipodal connected subset?
@Timkinsella Good. Make it an answer and I'll accept it. Thanks!
Jul
24
comment Surjectivity implies injectivity of finitely generated modules, localization?
@user26857 Right, but I only asked for a geometric meaning, not claimed that there should be. I asked since I feel that I don't really understand the determinant trick and it's still a very tricky stuff to me. I don't understand the underlying reason why viewing $M$ as an $A[X]$-module is that efficient, say, to obtain Frobenius normal forms for matrices, or in proof of Nakayama's lemma.
Jul
23
comment Surjectivity implies injectivity of finitely generated modules, localization?
@MartinBrandenburg Yeah, it might be frustrating. Incidentally, is there any geometric viewpoint of viewing $M$ as an $A[X]$ module by action $X.m=f(m)$?