A group homomorphism $\phi: G \rightarrow H$ induces isomorphisms w.r.t intersection and join between $\mathcal G$ and $\mathcal H$ I'm filling the gaps in the proof of Theorem 22 in textbook Algebra by Saunders MacLane and Garrett Birkhoff. Could you please verify if my second part. i.e. $\phi [S_{1} \lor S_{2}] = \phi [S_{1}] \lor \phi [S_{2}]$ is fine or contains logical mistakes?

Let $\phi: G \rightarrow H$ be a group homomorphism, $\mathcal G = \{S \le G \mid \operatorname{Ker} \phi \subseteq S \subseteq G\}$, and $\mathcal H := \{T \le H \mid \{1\} \subseteq T \subseteq \operatorname{Im} \phi\}$. For $S \le G$ and $T \le H$, the induced maps $\phi[\cdot]$ and $\phi^{-1}[\cdot]$ are defined by $\phi[S] := \{\phi(x) \mid x \in S\}$ and $\phi^{-1}[T] := \{x \in G \mid \phi(x) \in T\}$. Then $\phi[\cdot]$ and $\phi^{-1}[\cdot]$ are isomorphisms w.r.t intersection and join between $\mathcal G$ and $\mathcal H$.

First, we need a lemma to make the proof cleaner.

Lemma: If $S \in \mathcal G$, then $\phi^{-1}[\phi[S]] = S$.

Proof: Let $(x,y) \in S \times G$ such that $\phi(x) = \phi(y)$. Then $\phi (x y^{-1}) = \phi(x) \phi(y)^{-1} =1$ and thus $xy^{-1} \in \operatorname{Ker} \phi \subseteq S$. Hence $x y^{-1} \in S$ and thus $y \in S$. The result then follows.

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*$\phi [S_{1} \cap S_{2}] = \phi [S_{1}] \cap \phi [S_{2}]$
Clearly, $\phi [S_{1} \cap S_{2}] \subseteq \phi [S_{1}] \cap \phi [S_{2}]$. We have $\phi [S_{1}] \cap \phi [S_{2}] \subseteq \phi [S_{1}]$ and thus $\phi^{-1}[\phi [S_{1}] \cap \phi [S_{2}]] \subseteq \phi^{-1}[\phi [S_{1}]] \overset{(\star)}{=}S_1$. Similarly, $\phi^{-1}[\phi [S_{1}] \cap \phi [S_{2}]] \subseteq S_2$. Hence $\phi^{-1}[\phi [S_{1}] \cap \phi [S_{2}]] \subseteq S_1 \cap S_2$ and thus $\phi [S_{1}] \cap \phi [S_{2}] \subseteq \phi[S_1 \cap S_2]$.
$(\star)$: This is due to the lemma.


*$\phi [S_{1} \lor S_{2}] = \phi [S_{1}] \lor \phi [S_{2}]$
Notice that $S_1 \lor S_2 = \{s_1\cdots s_n \mid n \in \mathbb N^\times \text{ and } s_i \in S_1 \cup S_2\}$. Then $\phi [S_{1} \lor S_{2}] = \{ \phi(s_1)\cdots \phi(s_n) \mid n \in \mathbb N^\times \text{ and } s_i \in S_1 \cup S_2\}$. It follows from the lemma that $s_i \in S_1 \cup S_2 \iff \phi(s_i) \in \phi[S_1 \cup S_2]$. Hence $\phi [S_{1} \lor S_{2}] = \{h_1\cdots h_n\mid n \in \mathbb N^\times \text{ and } h_i \in \phi[S_1 \cup S_2]\}$. On the other hand, $\phi[S_1 \cup S_2] = \phi[S_1] \cup \phi[S_2]$ and consequently $\phi [S_{1} \lor S_{2}] = \{h_1\cdots h_n\mid n \in \mathbb N^\times \text{ and } h_i \in \phi[S_1] \cup \phi[S_2]\} = \phi [S_{1}] \lor \phi [S_{2}]$.


*$\phi[\cdot]$ is bijective

Let $S_1,S_2 \in \mathcal G$ such that $\phi[S_1] = \phi[S_2]$. By our lemma, $S_1 = \phi^{-1}[\phi[S_1]] = \phi^{-1}[\phi[S_2]] = S_2$. Then $\phi$ is injective. It follows from $\phi [\phi^{-1}[T]] = T \in \mathcal H$ that $\phi$ is surjective.
Similarly, $\phi^{-1} [\cdot]$ is bijective.
 A: Yes, this is fine, but it feels you have rather overegged it. Everything here follows from the straightforward properties of union and intersection of sets, with one application of the homomorphism property (which is hidden in your assertion that $\phi[S_1\vee S_2]=\{\phi(s_1)\dots\phi(s_n)\}$). I appreciate it's useful to write out more rather than less when we are first getting our heads around something, but I would try to strip this back as it unobfuscates (deobfuscates?) what's going on and why something is true.
As an example of what I mean, your entire proof of 1. is reduced to a single assertion (as it should be!) when it reoccurs for union in your proof of 2., without invoking the lemma or $\phi^{-1}$ or anything!
A: Proof of the lemma. The inclusion $S\subseteq \phi^{-1}[\phi[S]]$ is true for any map. Suppose $x\in\phi^{-1}[\phi[S]]$. Then $\phi(x)=\phi(y)$, for some $y\in S$. Therefore $xy^{-1}\in\ker\phi\subseteq S$. Hence $x=(xy^{-1})y\in S$.
Comment: you're hiding what you want to prove.

You're missing also the proof that the map $\mathcal{G}\to\mathcal{H}$ is well defined, but I guess it has already been proved that $\phi[S]$ is a subgroup of $H$, whenever $\phi\colon G\to H$ is a homomorphism and $S$ is a subgroup of $G$.

Proof that $\phi[S\cap T]=\phi[S]\cap\phi[T]$ (using different letters reduces clutter).
One inclusion is true for any map and any choice of subsets of the domain. Suppose $z\in\phi[S]\cap\phi[T]$. Then $z=\phi(x)=\phi(y)$, for some $x\in S$, $y\in T$. In particular $xy^{-1}\in\ker\phi$, so $xy^{-1}\in T$. Hence $x=xy^{-1}y\in T$, so $x\in S\cap T$ and $z=\phi(x)\in\phi[S\cap T]$.
Comment: you need not show that $y\in S\cap T$.

Proof that $\phi[S\vee T]=\phi[S]\vee\phi[T]$.
Since $S\subseteq S\vee T$, we have $\phi[S]\subseteq\phi[S\vee T]$ and similarly for $\phi[T]$. Therefore $\phi[S]\vee\phi[T]\subseteq\phi[S\vee T]$.
The reverse inclusion is easier if you show directly that a set of generators of $\phi[S\vee T]$ is $\phi[S]\cup\phi[T]$. Indeed, the latter set is a subset of the former; an element of $\phi[S\vee T]$ is an element of the form $\phi(a_1a_2\dotsm a_n)=\phi(a_1)\phi(a_2)\dotsm\phi(a_n)$, where $a_i\in S\cup T$ and the result follows.

Bijectivity. The lemma proves that $\phi^{-1}[\cdot]$ is the identity. Therefore $\phi[\cdot]$ is injective. But if $T\in\mathcal{H}$, we have $T=\phi[\phi^{-1}[T]]$, because this is true for any map and any subset of the range thereof. Now just notice that $\phi^{-1}[T]\in\mathcal{G}$.
Comment: you're missing the proof that $\phi[\cdot]$ is surjective.
