近世代数笔记-GTM73-Intro

年轻人的第一篇纯英notes，内容比较简单，主要是学习数学notes的英语表达QwQ

（其实是一些离散数学相关的前置知识）

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Let $f:A\rightarrow B$ and $g:B\rightarrow C$ be two mappings.

$f$ and $g$ injective $\implies$ $gf$ is injective; (9)
$f$ and $g$ surjective $\implies$ $gf$ is surjective; (10)
$gf$ injective $\implies$ $f$ is injective; (11)
$gf$ surjective $\implies$ $g$​ is surjective. (12)

Solution.

(9) For all $a$, $a'\in A$, $a≠a'\implies f(a)≠f(a')$

Let $b=f(a)$, $b'=f(a')$, and it follows from the definition of injectivity that $g(b)≠g(b')$, i.e. $g(f(a))≠g(f(a’)) $ . $\square$

(10) is similar to (9), omitted.

(11) Suppose that $f$​ is not injective. Then there exist $a$​, $a'\in A$​ with $a≠a'$​ and $f(a)=f(a')$​. So we have $g(f(a))=g(f(a'))$​, which contradicts the definition. Therefore, $f$​ is injective. $\square$​​​​

(12) is similar to (10), omitted.