Mathematics

The Yoneda Lemma The Yoneda lemma has been being complainted as “hard to understand” for a long time. Indeed it is not very easy, so I was trying to come up with a way to understand it more intuitively. Everyone knows the analogy by Ravi Vakil: You work at a particle accelerator. You want to understand some particle. All you can do are throw other particles at it and see what happens. If you understand how your mystery particle responds to all possible test particles at all possible test energies, then you know everything there is to know about your mystery particle. ...

Homology sphere A homology sphere is a (closed connected oriented) $n$-mainfolds with the same homology as the $n$-sphere. Poincaré first conjectured that any $n$-manifolds homologous (i.e. having the same homology) to the $n$-sphere should be homeomorphic to the $n$-sphere, then later he found a counterexample, which led him to a modified conjecture (which is the Poincaré’s conjecture). The counterexample he found is very interesting, one of the way to construct it is the folowing: ...

Using fields to detect integral homology? So in the universal coefficient theorem, by taking $R = \mathbb{Z}$, we know that there is a short exact sequence (which splits): $0 \rightarrow H_n(C_{*}) \otimes_R M \rightarrow H_n(C_{*} \otimes_R M) \rightarrow Tor^R_1(H_{n-1}(C_*), M) \rightarrow 0$ Some interesting cases are when $M = \mathbb{Z}_{p} \text{ or } \mathbb{Q}$ , do the homology over $\mathbb{Q}$ and $\mathbb{Z}_p$ tells us something about the integral homology? In particular, if $\tilde{H}(X; \mathbb{Q}) = 0$ and $\tilde{H}(X; \mathbb{Z}_p) = 0$ for all $p$, does it imply that $\tilde{H}(X) = 0$ ? .This is a problem suggested in [Kup2020]1. ...

Stone–Čech compactification One of the proof of the Stone–Čech compactification is to consider the continuous functions from a given topological space $X$ to $[0, 1]$. Crux 1: constructing $[0, 1]^{C}$ Let $C$ be the space of all continuous functions from $X$ to $[0, 1]$, consider $[0, 1]^{C}$, there is a natural map from $X$ to $[0, 1]^{C}$: for each $x \in X$, define $\phi(x) = f \mapsto f(x)$. With product topology $\phi$ is continuous. Also by Tychonnof’s theorem $[0, 1]^{C}$ is compact. Now the closure of the image is obviously compact Hausdorff, and it satisfies a nice property: any continuous map $f : X \to [0, 1]$ extends uniquely to a continuous map $\overline{\phi(X)} \to [0,1]$: for any $\phi(x)$, it has to map to $f(x)$! ...

I was going through a problem in Lee’s Topological Manifolds book1 (problem 7-12). It states that the infinite broom has a strong deformation retract to $(0, 0)$ but not for $(1, 0)$ (only a deformation retract). Below is some of my playful observations inspired by this problem. (Not restricted to the infinite broom space.) Inifinite Broom Note that the deformation retract to $(0, 0)$ itselfs imply the deformation retract to $(1, 0)$, since one can define the deformation retract to $(1, 0)$ by “first retract to $(0, 0)$, then push it to the point $(1, 0)$ along the line joining these two points”. ...

I am reviewing some fundamental algebra, and I just learnt something beautifully suggesting the connection of ring theory and geometry, which made me eager to learn some algebraic geometry (which I was not very interested when I was in undergrad). It is an exercise in Aluffi’s book1. Lets’ jump into it. The problem Let $K$ be a compact topological space, and let $R$ be the ring of continuous real-valued functions on $K$, with addition and multiplication defined pointwise. (i) For $p \in K$, let $M_p = \{f \in R | f(p) = 0\}$. Prove that $M_p$ is a maximal ideal in $R$. (ii) Prove that if $f_1, \dots , f_r \in R$ have no common zeros, then $(f1, \dots , fr)$ = $(1)$. (Hint: Consider $f^2_1 + \dots + f^2_r$ .) (iii) Prove that every maximal ideal $M$ in $R$ is of the form $M_p$ for some $p \in K$. (Hint: You will use the compactness of $K$ and (ii).) ...

Warning ...

Problem Let $I_{m} = \int^{2\pi}_{0} \cos(x)\cos(2x)\dots\cos(mx) dx$. For $m$ in $1, 2, \dots, 10$, for which $m$ is $I_m \neq 0$ ? Solution $I_m$ is non-zero if and only if $m \equiv 0, 3 \pmod{4}$. Main idea The official(seemingly? It is in the putnam problem book and a few solutions I found online do the same.) solution is to substitute $\cos x = \frac{e^{ix} + e^{-ix}}{2}$ followed by grouping the terms into $\cos x \cos (2x) \dots \cos (mx) = e^{\text{something}}$, and analyze the something. This approach of cause is of cause good in its own right: it is exactly how you split up the $e^{in\theta}$ terms into triginometric polynomials (or ) in Fourier analysis 101. However it is a little bit annoying to manipulate those powers of $e$. And for this problem it could be done much in a much simpler and elementary way: exploiting the point of symmetry. ...

Counting binary trees I was thinking about of the problem of balancing a binary tree, and my mind stumbled across to the question “How many different binary trees with labelled nodes can you make without changing the traversal order?”. After figuring out the answer myself I realized that the numbers of such binary trees are just Catalan numebrs (I was not really into combinatorics - now I am). For example, the following has traversal order 1->2->3->4->5->6->7 ...