I'm reading Theorem 22 in textbook Algebra by Saunders MacLane and Garrett Birkhoff.
It follows that $\phi_*S = \phi[S] := \{\phi(x) \mid x \in S\}$ and $\phi^*T = \phi^{-1}[T] := \{x \in G \mid \phi(x) \in T\}$.
and its proof
Here is Proposition 10:
Due to the properties of the set-valued functions $\phi[\cdot], \phi^{-1}[\cdot]$ induced from $\phi (\cdot)$, we always have $\phi_{*}\left(S_{1} \cap S_{2}\right) \subseteq \phi_{*} S_{1} \cap \phi_{*} S_{2}$. One sufficient condition for the equality to hold is that $\phi$ is injective.
Could you please elaborate on how Proposition 10 lead to the equality?
Best Answer
Clearly, $\phi [S_{1} \cap S_{2}] \subseteq \phi [S_{1}] \cap \phi [S_{2}]$. Below is my use of kernel to obtain $\phi [S_{1}] \cap \phi [S_{2}] \subseteq \phi [S_{1} \cap S_{2}]$:
Here is a lemma suggested by @user1729.
Proof: Notice that $\phi^{-1}[\phi[S]] = \{x \in G \mid \exists y\in S: \phi(x) = \phi (y)\} \overset{(\star)}{=} \{x \in G \mid \exists y\in S: xy^{-1} \in \operatorname{ker} \phi\}$. It follows that $\phi^{-1}[\phi[S]] = S \operatorname{ker} \phi$. Notice that $S \subseteq\phi^{-1}[\phi[S]]$. With similar reasoning in my above approach, we get $(x,y) \in S \times G$ and $\phi(x) = \phi(y)$ implies $y \in S$. Hence $\phi^{-1}[\phi[S]] = S$.
$(\star)$: This is because $\phi$ is a morphism of groups.
Then we use this lemma to obtain the latter inclusion as follows:
We have $\phi [S_{1}] \cap \phi [S_{2}] \subseteq \phi [S_{1}]$ and thus $\phi^{-1}[\phi [S_{1}] \cap \phi [S_{2}]] \subseteq \phi^{-1}[\phi [S_{1}]] \color{red}{=} S_1$. 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 $\phi [S_{1}] \cap \phi [S_{2}] \subseteq \phi[S_1 \cap S_2]$.