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Let $f:R \to R'$ be a homomorphism and $a\in R, a' \in R'$ such that $f(a)=a'$. Show that $\{x \in R | f(x) = a'\} = a + Ker(f)$.

I have tried find Ker(f) and stuck at $Ker(f) = \{x\in R | f(x) = 0_{R'}\} = \{x \in R | a' = 0_{R'}\}$.

What's next? How to proves above? Any idea? Thanks in advanced.

lap lapan
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1 Answers1

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If $f(x)=a'$, then $$f(x-a)=f(x)-f(a)=0,$$so $x-a\in\ker f$. Conversely, if $x=a+z$ with $f(z)=0$, then $f(x)=a'$.

Martin Argerami
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  • $f(x) = a' \Leftrightarrow f(x) = f(a) \Leftrightarrow f(x-a) = f(x) - f(a) = 0$ and then $x-a \in Ker(f)$. But, the problem is $\subseteq a+Ker(f)$, not only $Ker(f)$. How is it? – lap lapan Aug 22 '20 at 14:39
  • For the converse, from $x=a+z$ to $f(x) = a'$ is like this, right? : $x=a+z \Leftrightarrow f(x) = f(a+z) \Leftrightarrow f(x) = f(a) + f(z) \Leftrightarrow f(x) = a'$. – lap lapan Aug 22 '20 at 14:42
  • They are not "if and only if". But you only need implications. – Martin Argerami Aug 22 '20 at 15:51