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Components and Partitions1. PartitionsLet $X$ be a space. A partition of $X$ is the collection of disjoint subspaces of $X$ such that the union equals $X$. Example: $[0,1]$ and $(1,2]$ are partitions of $[0,2]$.Example: $[0,1]$ and $[1,2]$ are not partitions because they are not disjoint.2. ComponentsComponents are equivalence classes by defining the equivalence relation $\sim$ so that $x\sim y$..

The product of all primitive roots modulo $m$ is either 1 or -1.If $m=2$, then the product of all primitive roots is 1.If $r$ is a primitive root, then the inverse $\bar{r}$ is also a primitive root.When calculating the product of all primitive roots, we can pair each one with its inverse, so that the product of each pair becomes 1. Now we are left with one element(not known if it exists or not)..

If integer $m>2$ has a primitive root and $x^2\equiv1\ (\text{mod} \ m)$, then either $x\equiv1\ (\text{mod} \ m)$ or $x\equiv-1\ (\text{mod} \ m)$.Proof.Solve the congruence $x^2\equiv1\ (\text{mod} \ m)$. It is equivalent to solving the congruence $2y\equiv\text{ind}_r(1)\equiv0\ (\text{mod} \ \phi(m))$Since $\text{gcd}(2,\phi(m))=2$ (because $\phi(m)$ is even), this congruence has 2 solutions..

There are several ways to prove this. In my last article, I introduced you the proof as a corollary, but this only works if $m$ has a primitive root. https://wangdo.tistory.com/85 2 be an integer with a primitive root r. Then, r^(ϕ(m)/2) ≡ −1 (mod m)" data-og-description="Let $m>2$ be an integer with a primitive root $r$. Then, $r^{\phi(m)/2}\equiv-1 \ (\text{mod} \ m)$.Proof.$\text{gcd}(-1,m)=1..

Let $m>2$ be an integer with a primitive root $r$. Then, $r^{\phi(m)/2}\equiv-1 \ (\text{mod} \ m)$.Proof.$\text{gcd}(-1,m)=1$. Therefore, there exists an integer $02$. Therefore, $2i=\phi(m)$. * Here we can obtain the corollary that $\phi(m)$ is an even number if $m>2$. (Note that $\phi(2)=1$ is odd)We get $i=\phi(m)/2$. End of proof.

위상수학 공부하다가 bounded의 정의가 모호하다는 것을 발견했다.인터넷에서 흥미로운 글을 발견했다. 참고: https://math.stackexchange.com/questions/1500897/bounded-set-in-metric-vs-ordered-spaces The term "bounded" is overloaded. 여기서 overloaded는 과적했다는 뜻인데, 컴퓨터 공부한 사람들은 한 함수가 여러 가지 방법으로 정의되었다는 걸 뜻한다고 알고 있을 거다.이 문장에서 overloaded 역시 그런 뜻인데 용어 하나에 여러가지 뜻이 담겨 있고 문맥에 따라 어느 뜻으로 쓰였는지 판단할 수 있다는 거다. 다음은 bounded이라는 용어에 대한 나의 설명.1. In METRIC TOPOLOGY, ..

Munkres 54.8 Let $p:E\rightarrow B$ be a covering map. If $B$ is simply connected and $E$ is path connected, then $p$ is a homeomorphism.Proof.Let $x_1, x_2\in E$ and $\tilde{f}$ be a path from $x_1$ to $x_2$. Then $f=p\circ\tilde{f}$ is a path from $p(x_1)$ to $p(x_2)$ and by the path lifting theorem, $\tilde{f}$ is the unique path lifting of $f$. Now, assume $p(x_1)=p(x_2)$, then $f$ is a loop..

Munkres 53.6(b)Let $p:E\rightarrow B$ be a covering map. If $B$ is compact and $p^{-1}(b)$ is finite for all $b\in B$, then $E$ is compact.Proof.Let $\{U_\alpha\}$ be an open cover of $E$. *Recall: a covering map is an open map.If $E$ is an $n$-fold covering of $B$, $p^{-1}(b)=\{e_1, e_2,...,e_n\}$, there exists a finite subset of $\{U_\alpha\}$ that covers $p^{-1}(b)$, say $\{U_1, U_2, ...,U_n\..

Let $U\subset B$ be evenly covered by $p:E\rightarrow B$ and $\{V_\alpha\}$ be the partition of $p^{-1}(U)$ into pieces. Number of elements of $\{V_\alpha\}$ is equal to the number of elements of $p^{-1}(b)$ for all $b\in U$.Because each of $p|_{V_\alpha}$ is a homeomorphism, exactly one element of $p(b)$ exists in $V_\alpha$. No more, no less.

Proposition. Let $p:E\rightarrow B$ a covering map. If $U\subset B$ is evenly covered by p and $U'\subset U$ is open, $U'$ is evenly covered by p.Proof.Let $\{V_\alpha\}$ be the partition of $p^{-1}(U)$ into pieces and $p|_{V_\alpha}:V_\alpha\rightarrow U$, a homeomorphism. Let $V_\alpha=(p|_{V_\alpha})^{-1}(U)$ are disjoint, so $(p|_{V_\alpha})^{-1}(U')$ are disjoint and $\bigcup (p|_{V_\alpha..