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My textbook states the following proposition: Let $f:R \rightarrow S$ be a ring homomorphism and let $s$ be in the image of $f$. Then $\{r \in R \mid f(r) = s\}$ is in one-to-one correspondence with $\ker(f)$.

What does it mean to have one to one correspondence with $\ker(f)$? Does it mean the set $\{r \in R \mid f(r) = s\}$ and the set $\ker(f)$ have the same cardinality?

Thanks!

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    "One-to-one correspondence" is a poor (in my opinion, anyway) way of saying "bijection."2011-04-01

2 Answers 2

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HINT $\rm\ $ If $\rm\:f(r)\ =\ s\ $ then \rm\ f(r')\ =\ s\ \iff\ 0\ =\ f(r')-f(r)\ =\ f(r'-r)

Hence $\rm\ \ f^{-1}(s)\ =\ r\ +\ ker\ f\ =\: $ particular + homogeneous solution, as in linear algebra.

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    @mat That's a more general definition of kernel. For rings the above shows that the general kernel is determined by $[0]$, the equivalence class of $\:0$, i.e. $\rm\ (r',r) \in ker\ f\ $ $\iff$ $\rm\:f(r')=f(r)\:$ $\iff$ $\rm r'-r\in f^{-1}(0) = ker_0\:f\:.$ But perhaps it is better to ignore this more general notion at first. Note: above I didn't mean "equalizer" but rather fiber, or preimage.2011-04-02
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It means that there is a map $g: \{r\in R | f(r) = s\} \rightarrow \operatorname{ker}f$ such that every element of $\operatorname{ker}f$ is the image of one and only one element of $\{r\in R | f(r) = s\}$. This implies that the cardinality is the same.

In this case, if s'\in R is such that f(s') = s, then one such map is g(r) = r - s'.

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    think of a group homomorphism $f:G\to H$, then $G/\text{ker}f\cong\text{im}f$, ie everything in a coset gets mapped to the same element...2011-04-01