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Here is the part of the book (user's guide to spectral sequences) I am reading:

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I have the following questions about this part:

1- Why we are sure that $k$ has an inverse (the user used $k^{-1}$)?

2- if $d = j \circ k$ why $\ker d = k^{-1}(\ker j)$?

3- Why $k (\ker d) = k'(\ker d/ \operatorname{im} d)$?

4- Why $ k'(\ker d/ \operatorname{im} d)$ is considered as the whole image of $k'$?

Could anyone help me answer those questions please?

Ѕᴀᴀᴅ
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Intuition
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    The exponent $-1$ is used to denote the inverse image. Note that $\ker(j \circ k) = (j \circ k)^{-1}({0}) = k^{-1}(j^{-1}({0})) = k^{-1}(\ker j)$ – azif00 Apr 25 '23 at 22:46
  • @azif00 so it is not necessarily has an inverse in our case here, am I correct? – Intuition Apr 25 '23 at 23:41
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    @Secretly yes there isn't necessarily an inverse. If A is a set then $k^{-1}(A)$ is defined as follows: $k^{-1}(A)={x\in dom(k)|k(x)=a,,for,some,a\in A}$, where by $dom(k)$ I denote the domain of $k$. So $k^{-1}(A)$ is again a set rather than the inverse of the function $k$. – Fotis Apr 26 '23 at 00:14
  • Got it, thanks. @Fotis how about the rest of my questions? – Intuition Apr 26 '23 at 00:15
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    @Secretly regarding question 4, $k'$ is a map $E'\rightarrow D'$. So we get $k'(E')=im(k').$ But $E'$ is defined as: $E'=kerd/imd$, so the desired equality follows – Fotis Apr 26 '23 at 00:44
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    Question 3 should also follow directly by the way $k'$ is defined, since $k'(e+im(d)):= k(e)$ – Fotis Apr 26 '23 at 00:57
  • @Fotis I am trying to digest your answer to question 3 now. – Intuition Apr 26 '23 at 08:43

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