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Let $f\colon X\to Y$ be a function. Let $(A_{i})_{{i\in I}}$ be a family of subsets of $X$, and let $(B_{j})_{{j\in J}}$ be a family of subsets of $Y$, where $I$ and $J$ are nonempty index sets.
Then, it is easy to prove, directly from definitions, that the following hold:

$f(\bigcap\limits_{{i\in I}}{A_{i}})\subseteq\bigcap\limits_{{i\in I}}{f(A_{i})}$ (i.e., the image of an intersection is contained in the intersection of the images)

$A\subseteq f^{{1}}(f(A))$ for any $A\subseteq X$ (where $f^{{1}}(f(A))$ is the inverse image of $f(A)$)

$f(f^{{1}}(B))\subseteq B$ for any $B\subseteq Y$

$f^{{1}}(Y\setminus B)=X\setminus f^{{1}}(B)$ for any $B\subseteq Y$

$f^{{1}}(\bigcup\limits_{{j\in J}}{B_{j}})=\bigcup\limits_{{j\in J}}{f^{{1}}(% B_{j})}$ (the inverse image of a union is the union of the inverse images)

$f^{{1}}(\bigcap\limits_{{j\in J}}{B_{j}})=\bigcap\limits_{{j\in J}}{f^{{1}}(% B_{j})}$ (the inverse image of an intersection is the intersection of the inverse images)

$f(f^{{1}}(B))=B$ for every $B\subseteq Y$ if and only if $f$ is surjective.
For more properties related specifically to inverse images, see the inverse image entry.
Further, the following conditions are equivalent (for more, see the entry on injective functions):

$f$ is injective

$f(S\cap T)=f(S)\cap f(T)$ for all $S,T\subseteq X$

$f^{{1}}(f(S))=S$ for all $S\subseteq X$

$f(S)\cap f(T)=\varnothing$ for all $S,T\subseteq X$ such that $S\cap T=\varnothing$

$f(S\setminus T)=f(S)\setminus f(T)$ for all $S,T\subseteq X$
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