It will, and I'm thinking if anything you don't want them to be distinct
Since otherwise you may not get the nice inclusion I mentioned above. If you have a point in $SP^n(X)$ for which one of the coordinates is already the basepoint, how are you gonna fit that in?
@Eric I don't know anything else offhand
But I'm only just starting, originally I was gonna do this paper in the break but Peter was like "Actually how about we put a deadline?"
Wait how do you otherwise prove that $K(\mathbb{Z},2) = \mathbb{CP}^{\infty}$?
@EricSilva One of the topologists in my math department said using homology to prove invariance of domain was like bringing a nuclear bomb to a thumb war
Invariance of domain is a theorem in topology about homeomorphic subsets of Euclidean space Rn. It states:
If U is an open subset of Rn and f : U → Rn is an injective continuous map, then V = f(U) is open and f is a homeomorphism between U and V.
The theorem and its proof are due to L. E. J. Brouwer, published in 1912. The proof uses tools of algebraic topology, notably the Brouwer fixed point theorem.
== Notes ==
The conclusion of the theorem can equivalently be formulated as: "f is an open map".
Normally, to check that f is a homeomorphism, one would have to verify that both f and its inverse...
@Daminark You use it to show that a topological manifold $M$ cannot be both of dimension $n$ and $m$ for $n \neq m$ if I recall correctly
Wait why does this give you that? Do you show that all open sets in $\mathbb{R}^n$ are homeomorphic and then given two different dimensions find sets which aren't or something?
Nah in the wiki article it says - "An important consequence of the domain invariance theorem is that $\mathbb{R}^n$ cannot be homeomorphic to $\mathbb{R}^m$ if $m \neq n$. Indeed, no non-empty open subset of $\mathbb{R}^n$ can be homeomorphic to any open subset of $\mathbb{R}^m$ in this case."
@Daminark Here's a simpler proof. Suppose $M$ is both a $n$ and $m$-dimensional manifold for $n \neq m$. Pick a point $p \in M$, there exists two charts $(U, \phi)$ and $(V \varphi)$ around $p$ for which $\phi$ maps $U$ homeomorphically onto $\widetilde{U} \subseteq \mathbb{R}^n$ and $\varphi$ maps $V$ homeomorphically onto $\hat{V} \subseteq \mathbb{R}^m$., by definition of $M$ both being a $n$ and $m$-dimensional manifold
Put $W = U \cap V$, we have that $W$ is open and $\phi|_{W} : W \to \widetilde{W} \subseteq \mathbb{R}^n$ and $\varphi|_{W} : W \to \hat{W} \subseteq \mathbb{R}^m$. Both of which are homeomorphisms being the restrictions of homeomprhisms. Since $W \cong \widetilde{W}$ and $W \cong \hat{W}$, we have $\widetilde{W} \cong \hat{W}$, contradicting invariance of domain.
It can be like coffee or beer or scothch. At first it seems real dumb. but if you just tell yourself its amazing for a while, you start to believe it and notice the nuances
I wont mix single-malt stuff because there's just no point, IMO. When you mix it with a greater volume of coke, you cant tell the difference between a $20 bournbon and a $200 bourbon
@Daminark I too think your application is more underwhelming than theorem.
$\Bbb{CP}^\infty = K(\Bbb Z, 2)$ is actually just a computation
there's a fibration $S^1 \to S^\infty \to \Bbb{CP}^\infty$.
Or, if you want to be r/iamverysmart, $\Bbb{CP}^\infty = BU(1)$ because maps to that classifies complex line bundles/U(1) bundles. U(1) = BZ, so it's a $B^2\Bbb Z$
Like I either couldn't draw the picture of a situation at all, even sometimes in basic cases, or I could draw it and I couldn't figure out how to extract any information
And the topics in that class were classical mechanics, E&M, and then waves. First two quarters were hell because our problems were extremely physical/geometric
That I am bad at the things in between is also why I suck at physics
In general, I am bad at any things that has many parameters changing at the same time not in a vectorised manner
e.g. computing limits
By "vectorised manner", I mean there are computations that you can just do by treating the whole problem as an array or matrix problem
In set theory, replacement is like a vectorised computation to me because I am applying functions to all elements in a set at the same time, thus in a sense, there is really only ONE thing that is changing
but dynamics and analysis is different. Unless you have the ability to keep track of uncountably many paths, there is no way to know where you end up
and the worst of them is limits and epsilon delta proofs
(because you have to justify a given epsilon by going backwards)
I can understand the intuitive meaning of epsilon deltas and also can read the formal meaning, but I often fail to find the required epsilon because there is too many things happening at the same time
which is why during this catch up, one reason I want to do point set topology first and use it to bridge back to analysis is because if I can manage something that abstract and full of infinities, then there should not be trouble for me to deal with something as concrete as epsilon deltas
For example, reading Munkres plus one remark of Steamy and Alessandros does help me understand functions better
I briefly read about them, they are very nice to me because you can reason about topologies that are not first countable with them (and I have a tendency to do exercise on not very nice topologies to ensure my intuition being built is general enough), but I have not read very deeply into them yet. I do suspect they will help me to grasp limit computation though due to their general nature
More generally, I like pathological things not just because thy are weird, but because intuition gain on them can often be aplied to nice things easily
This is because pathological things often tell us why some things fail, and hence by contrapositive, why things are nice
I think suffice to say I sometimes find more trouble understanding some nice things compared to the pathological ones
That reminds me of what Peter Scholze said about p-adic analysis over ordinary analysis - it just looks more natural to him now, and he has to work to use the ordinary stuff mere mortals use.
Pretty sure that cannot happen for finite vs infinite, because we are mortals, and we always have one more way to comprehend the finite which we don't for infinite: Count them!
I am so confident on that claim that I am dared to say the following: Whoever mathematician that claimed "Now I find finite numbers much, much more confusing than infinite numbers." should deserve 5 Fields Medals
@AlessandroCodenotti So if $(X_i)_{i \in I}$ are a countable family of metrizable topological spaces, the product topology on $X =\prod_{i \in I}X_i$ is the same as the metric topology on $X$?
Suppose $(X,d)$ is metric space. I want to show that $A \subseteq X$ is open iff $A$ is union of open balls.
My attempt. suppose $A$ is open, then for every $x \in A$, there exists $r>0$ such that $B(x,r) \subset A$ by definition. We claim that $A = \bigcup_{x\in A} B(x,r) $. To see this, pick $...
I am working on exercise 1.1 and I think the way to do this would be to show that open sets are homeomorphic to $R^n$ or open balls in $R^n$. Is this even true? I'm not sure how to go about proving it.
btw, the exercise is from Lee's Smooth Manifolds book.
You're essentially saying that if points $a$ and $b$ are equidistant from $x$, then the line between $a$ and $b$ is perpendicular to the line between $x$ and $\frac{a+b}2$, their midpoint
That said, I think you could do it algebraically by expanding it with the distributive property
If $\widetilde{M} \to M$ is a universal covering of a smooth manifold, is it clear that the natural map $\pi_1(M) \backslash T\widetilde{M} \to TM$ is a diffeomorphism?
If I have a function $h$ which is a shifted version of another $g$ and I have a Fourier series approximation for $h$, what can I say about a Fourier series approximation for $g$?
I think I have to at least shift in an equivalent way (to the way $h$ is shifted from $g$) the Fourier series for $h$ in order to make it approximate $g$
@BalarkaSen Ah I see, and it's one-to-tone because $\widetilde{M} \to M$ is a local diffeomorphism hence its differential at any point is an isomorphism, right?
@Vrouvrou You can use the same trick to calculate that… but remember that if $\langle X,Y\rangle=0$ (that is, if $X$ and $Y$ are perpendicular) then $\|X\|^2+\|Y\|^2=\|X+Y\|^2$. This is the Pythagorean theorem. (You can prove this formulation of it by expanding $\|X+Y\|^2=\langle X+Y,X+Y\rangle$.)
@BalarkaSen I'm trying to see injectivity. If I am given vectors $v,w \in T_p\widetilde{M}$ with $d \pi_p(v) = d \pi_p(w)$, how do I see that $v$ and $w$ are in the same orbit of the action of $\pi_1M$ on $T_p\widetilde{M}$?
If there's no contradiction regarding a statement, then you can assume it is true whenever you want.
Once you find a contradiction in a logical argument that follows from that statement, you know it isn't true anymore, thus you can't assume it is true in your successive arguments.
Can some1 review the following question? https://math.stackexchange.com/questions/2549516/question-on-proof-regarding-unique-homomorphism-in-finitely-generated-vector-spa I am still wondering if I have to find a contradiction for i)one example ii)all vector spaces iii) every family of vectors in every vector space