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Claim: Let $\varphi:G \rightarrow H$ be a group homomorphism. Then the maps $A \mapsto \varphi(A)$ and $B \mapsto \varphi^{-1}(B)$ (where here we mean preimage) specify inverse bijections between the sets $\{A \leq G \mid \ker(\varphi) \subseteq A\}$ and $\{B \leq H \mid B \subseteq \text{Im}(\varphi)\}$.

Proof: Let $A \leq G$ contain the kernel, then it's evident that $A \subseteq \varphi^{-1}(\varphi(A))$. Let $g \in \varphi^{-1}(\varphi(A))$, then this means that $\varphi(g)= \varphi(a)$ for some $a \in A$. Then because $A$ is a subgroup $\varphi(ga^{-1}) = \varphi(g)\varphi(a^{-1}) = 1_H$ and so $ga^{-1} \in \ker(\varphi) \subseteq A$, but then because $A \leq G$ we have $(ga^{-1})a = g \in A$ and so the other inclusion holds.

The argument above isn't difficult, but I can't understand why we need to do this in the first place? If we're considering just images and preimages, doesn't it follow immediately from the definition that $(\varphi \circ \varphi^{-1})(A) = A =(\varphi^{-1} \circ \varphi)(A)$? What about the statement necessitates we prove it via equality of sets? Thanks in advance for the clarification.

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    $\begingroup$ Here $\phi^{-1}$ is not the inverse of $\phi$, but inverse image. For example, if you compute $\phi(\phi^{-1}(A))$ [let's not use the notation $\phi\circ\phi^{-1}$] that might not be all of $A$, for all functions $\phi$ and all subsets $A$. Similarly $\phi^{-1}(\phi(A))$ can be larger than $A$. $\endgroup$
    – plop
    Oct 5, 2022 at 20:58
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    $\begingroup$ For example, the function $f:\{0,1\}\to\{0,1\}$ sending $f(0)=f(1)=0$, would have, for $A=\{0,1\}$, $f(f^{-1}(A))=\{0\}$ and for $B=\{0\}$ $f^{-1}(f(B))=\{0,1\}$. $\endgroup$
    – plop
    Oct 5, 2022 at 21:03
  • $\begingroup$ Oh of course, my fault. Don't know why I didn't realize I was treating the preimage like a function. Thanks! $\endgroup$
    – Irving Rabin
    Oct 5, 2022 at 21:04

2 Answers 2

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  1. You only did half of the work: you didn't prove that forall subgroup $B$ in your second set, $\varphi(\varphi^{-1}(B))=B$.
  2. You did that first half well and (with the same argument) you could have proven that more generally $\forall X\subset G\quad\varphi^{-1}(\varphi(X))=(\ker\varphi)X$. Similarly, for the second "half of the work", you could easily prove more generally (and here, without using that $\varphi$ is a morphism) that $\forall Y\subset H\quad\varphi(\varphi^{-1}(Y))=Y\cap\operatorname{im}\varphi$.
  3. Using the two properties $\varphi^{-1}(\varphi(X))=(\ker\varphi)X$ and $\varphi(\varphi^{-1}(Y))=Y\cap\operatorname{im}\varphi$, it is easy to construct, for most morphisms $\varphi:G\to H$, a subgroup $A$ of $G$ and a subgroup $B$ of $H$ such that $A\subsetneq\varphi^{-1}(\varphi(A))$ and $\varphi(\varphi^{-1}(H))\subsetneq H$. Exercise: for which morphisms do such counterexamples exist?
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  • $\begingroup$ I understand that the preimage isn't a function so the notation I used is wrong. Would you be willing to clarify what's wrong with my understanding in the following statement? If we consider $\varphi(A)$ this is all $\varphi(x)$ such that $x \in A$, then if we take the preimage of $\varphi(A)$ we have all the elements in $G$ that $\varphi$ takes to $\varphi(A)$, is this not just $A$? That's all I would have written to exhibit the one-sided inverse here. $\endgroup$
    – Irving Rabin
    Oct 5, 2022 at 23:14
  • $\begingroup$ It is just $A$ if $\varphi$ is injective. But else it may be bigger, because there may be some $y\notin A$ having the same image as some $x\in A$. $\endgroup$
    – Anne Bauval
    Oct 5, 2022 at 23:17
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    $\begingroup$ Thanks so much! I hadn't realized I was tacitly assuming $\varphi$ was 1-1, really cleared things up! $\endgroup$
    – Irving Rabin
    Oct 5, 2022 at 23:21
  • $\begingroup$ Just to hammer the point home, to start the second part would you argue the following; $\varphi(\varphi^{-1}(B)) \subseteq B$ since the image of the preimage of $B$ is certainly contained in $B$ but may not be all of $B$ as $\varphi$ isn't necessarily surjective? Then let $b \in B$ and so on. $\endgroup$
    – Irving Rabin
    Oct 5, 2022 at 23:53
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    $\begingroup$ Yes ($\forall Y\subset H\quad\varphi(\varphi^{-1}(Y))=Y\cap\operatorname{im}\varphi$) $\endgroup$
    – Anne Bauval
    Oct 5, 2022 at 23:55
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It isn't true that $(\varphi^{-1} \circ \varphi)(A)=A$.
Suppose $G = \mathbb{Z},\ A = \{0\},\ H = \{0\}$ and $\varphi(n) = 0\quad \forall n \in \mathbb{Z}$
then $(\varphi^{-1} \circ \varphi)(A) = G \neq A$.

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    $\begingroup$ Beware: not $\varphi^{-1}\circ\cdots$ (see comments above) $\endgroup$
    – Anne Bauval
    Oct 5, 2022 at 21:08
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    $\begingroup$ Beware again: $\varphi^{-1}$ alone has no meaning, it is not a function, just a notation for the inverse image. Put $\circ$ nowhere. $\endgroup$
    – Anne Bauval
    Oct 5, 2022 at 21:16
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    $\begingroup$ Thanks @AnneBauval that was clarifying. $\endgroup$
    – Irving Rabin
    Oct 5, 2022 at 23:08

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