In mathematics, the disjoint union (or discriminated union) of a family
(
A
i
:
i
∈
I
)
{\displaystyle (A_{i}\colon i\in I)}
of sets is a set
A
=
⨆
i
∈
I
A
i
,
{\displaystyle A=\bigsqcup _{i\in I}A_{i},}
with an injective function of each...
is there any good reason to use "$A\subset B$" rather than "$A\subseteq B$" for subsets? (taking less effort to write does not count as a good reason for the purpose of this question)
Let $X\cup_f Y$ be an adjunction space. Let $q:X\coprod Y \rightarrow X\cup_f Y$ be the associated quotient map.
Where $\sim$ is generated by $a\sim f(a)$ for all $a$
Show that $q$ is injective.
My attempt:
Case 1: Let $(x_1,0),(x_2,0)$ ( I will write it as $x_1,x_2$) be in $X$. So $q(x_1)=q(...
A possible example: Consider $\exp(-1/x^2)$ for $x > 0$. Is there a precise sense in which $f(x) = \begin{cases} 0 & x = 0 \\ \exp(-1/x^2) & \text{otherwise} \end{cases}$ is the "most natural" extension to all of $\mathbb{R}$?
In the second question I linked to, Gammel refers to "a kind of analytic continuation which differs from the usual kind" which he attributes to Borel, but I haven't been able to determine what that is.
@geocalc33 The (modern) language of manifolds will be overwhelming at the start. That's a given. It's the subject where the definitions are hard, and the theorems are easy.
You can escape a lot of the notation if you do something like Guillemin and Pollack's differential topology book, but then the notation catches up to you later.
@user76284 No. When I'm not constructing bump functions, I'm equally fond of the function $f(x)=\begin{cases} e^{-1/x^2}, & x\ne 0 \\ 0, & x=0\end{cases}$.
Another example given by Shapiro in Generalized Analytic Continuation is this: Let $z_k$ be a sequence of points which is dense on the unit circumference and $c_k$ be complex numbers such that $c_k \neq 0$, $\sum_{n=0}^\infty |c_k| < \infty$. Then the function $f(z) = \sum_{n=0}^\infty \frac{c_k}{z-z_k}$ for $|z| < 1$ cannot be continued analytically across any point of the unit circle.
Because you get a dense set of poles on the unit circle.
"Nevertheless, it has nontangential limiting values almost everywhere on $|z|=1$, which coincide almost everywhere with the non-tangential limiting values of the analytic function defined by the same series for $|z| > 1$."
If $G$ is a finite group and all its irreducible representations have a non degenerate G-invariant symmetric bilinear form, $\sum_{\rho\in\hat G}dim(\rho)$ is the number of elements of order 2 of G
@TedShifrin I might be crazy, but say I have two orthogonal projections, $P,Q$ and $P(Q(x)) = x$. I want to show that $x\in \im P \cap \im Q$ but I am struggling to figure it out. >: ( any tips?
It's probably just some little thing I am missing, but I tried to go about looking at $\|P(Q(x))\|^2$ with the inner product, but to no avail
Like getting something like $\langle Q(P(Q(x))),x\rangle = \|x\|^2$ sort of thing.
Didn't seem helpful. I also tried to split it up into $x = v_1 + v_2 + k_1 + k_2$ where v_i are in the images of P and Q respectively and the k_i in their kernels (orthogonal complements).
Let $X\cup_f Y$ be an adjunction space. Let $q:X\coprod Y \rightarrow X\cup_f Y$ be the associated quotient map.
Where $\sim$ is generated by $a\sim f(a)$ for all $a$
Show that $q$ is injective.
My attempt:
Case 1: Let $(x_1,0),(x_2,0)$ ( I will write it as $x_1,x_2$) be in $X$. So $q(x_1)=q(...
