Proof that $f(\bigcap_{\lambda\in L} A_\lambda) = \bigcap_{\lambda \in L} f(A_\lambda)$ iff $f$ is injective I intend to show that $f\colon X\to Y$ is injective iff for any family $(A_{\lambda})_{\lambda\in L}$ of subsets of $X$ we have $f(\bigcap_{\lambda\in L} A_\lambda) = \bigcap_{\lambda \in L} f(A_\lambda)$.
First, let's prove that for $f\colon X\to Y$and $A\subset X$, $f$ is injective iff $A= f^{-1}(f(A))$
Proof. ($\Rightarrow$). By contradiction. If $f$ is injective, then for any $x_1,x_2\in X$ we have $x_1\neq x_2 \implies f(x_1)\neq f(x_2)$. Choose $x_1\in A\subseteq X$ and $x_2 \in X\setminus A$, it's obvious that $f(x_1)\in f(A)$ e $f(x_2)\in f(X\setminus A)$. Then every $y\in f(A)$, because $f$ is injective, corresponds to an unique point $x\in A\subseteq X$ such that $f(x)=y$. Then for all $A\subseteq X$ we can say that $f^{-1}(f(A))=A$. If there was an element $x_3\in X\setminus A$ such that $f(x_3)\in f(A)$, then $f$ wouldn't be injective because $f(x_3)\in f(A)$, exactly because it is located in $f(A)$, would have to correspond to some $x_3\in A\subseteq X$. Then the injectivity of $f$ implies that for all $A\subseteq X$ we have $f^{-1}(f(A))=A$.
($\Leftarrow$). By contradiction. Consider $x_1\in A\subseteq X$ and $x_2\in X\setminus A$. Then $f(x_1)\in f(A)$ and $f(x_2)\not\in f(A)$. We have that $f^{-1}(f(A)) = A$, then every element in $f(A)$ corresponds to some element in $A\subseteq X$. Consider that $f$ is not injective, i.e., the case that $f(x_1)=f(x_2)\in f(A)$ and $x_1\neq x_2$. We would have then $f^{-1}(f(A))\neq A$, because $x_2\not\in A$. Then, $f$ is injective.
Now to the main proof.
Proof. Direct proof that $f$ is injective $\iff f(\bigcap_{\lambda\in L} A_\lambda) = \bigcap_{\lambda \in L} f(A_\lambda)$.
\begin{align*}
    f(x)\in f(\bigcap_{\lambda\in L} A_\lambda) &\iff x\in f^{-1}(f(\bigcap_{\lambda\in L} A_\lambda))\\
    &\iff x\in \bigcap_{\lambda\in L} A_\lambda \ \text{because $f$ is injective}\\
    &\iff x\in A_\lambda, \text{for all} \ \lambda\in L\\
    &\iff f(x) \in f(A_\lambda), \text{for all} \ \lambda\in L\\
    &\iff f(x) \in \bigcap_{\lambda\in L} f(A_\lambda)
\end{align*}
Can I do this?
I couldn't show that if $f(\bigcap_{\lambda\in L} A_\lambda) = \bigcap_{\lambda \in L} f(A_\lambda)$ then $f$ is injective without using the fact that $f(\bigcap_{\lambda\in L} A_\lambda) = \bigcap_{\lambda \in L} f(A_\lambda)$ only stands if and only if $f^{-1}(f(\bigcap_{\lambda\in L} A_\lambda)) = \bigcap_{\lambda\in L} A_\lambda$, and this is only true if and only if $f$ is injective.
 A: A much simpler argument is possible. It’s easy to see that it’s always true that
$$f\left[\bigcap_{\lambda\in\Lambda}A_\lambda\right]\subseteq\bigcap_{\lambda\in\Lambda}f[A_\lambda]$$
for any family $\{A_\lambda:\lambda\in\Lambda\}$ of subsets of $X$. Suppose that $f$ is injective, and let $x\in\bigcap_{\lambda\in\Lambda}f[A_\lambda]$. Then for each $\lambda\in\Lambda$ there is an $a_\lambda\in A_\lambda$ such that $x=f(a_\lambda)$. Fix $\lambda_0\in\Lambda$; since $f$ is injective, $a_\lambda=a_{\lambda_0}$ for each $\lambda\in\Lambda$, so $a_{\lambda_0}\in\bigcap_{\lambda\in\Lambda}A_\lambda$, and $x=f(a_{\lambda_0})\in f\left[\bigcap_{\lambda\in\Lambda}A_\lambda\right]$. Thus,
$$\bigcap_{\lambda\in\Lambda}f[A_\lambda]\subseteq f\left[\bigcap_{\lambda\in\Lambda}A_\lambda\right]$$
and hence
$$f\left[\bigcap_{\lambda\in\Lambda}A_\lambda\right]=\bigcap_{\lambda\in\Lambda}f[A_\lambda]\,.$$
Now suppose that $f$ is not injective. Then there are $x,y\in X$ such that $x\ne y$, and $f(x)=f(y)$. Let $A_0=\{x\}$ and $A_1=\{y\}$; then
$$\varnothing=f[\varnothing]=f[A_0\cap A_1]\subsetneqq f[A_0]\cap f[A_1]=\{f(x)\}\,.$$
