Prove that $\bigcap f(A_{\alpha}) \subseteq f(\bigcap A_{\alpha} ) $

Question

Let $\ f: X \rightarrow Y $ an injective function and $\ \{A_{\alpha}\}_{\alpha \in I} $ a group of subgroups of $\ X $

prove:

$$\ \bigcap_{\alpha \in I} f \bigl( A_{\alpha} \bigl)\subseteq f \Bigl( \bigcap_{\alpha \in I} A_{\alpha} \Bigl) $$

my attempt:

$\ w_1 \in \bigcap_{\alpha \in I} f(A_{\alpha}) \Rightarrow w_1 \in f(A_{\alpha}) $ for every $\ \alpha \in I $ then there is a $\ w_2 \in X $ such that $\ f(w_2) = w_1 $

but now I need to prove that this $\ w_2 $ is in $\ \bigcap_{\alpha \in I } A_{\alpha} $ so that I can claim $\ w_1 \in f(\bigcap_{\alpha \in I} A_{\alpha}) $

but I don't know how to prove it?

Answer 1

Fix $\alpha$. Then $w_1 \in f(A_\alpha)$ implies there is $w_2 \in A_\alpha$ such that $f(w_2) = w_1$.

Fix another $\alpha'$. We know $w_1 \in f(A_{\alpha'})$ as well, so there exists $w'_2 \in A_{\alpha'}$ such that $f(w'_2) = w_1$.

What does injectivity of $f$ tell you about $w_2$ and $w'_2$?