Prove $B_1 \subseteq B_2 \Rightarrow f^{-1}(B_1) \subseteq f^{-1}(B_2)$ > Let $f : X \rightarrow Y$ and let $B \subseteq Y$. The inverse image of $B$ is a subset of $X$ defined by $$f^{-1}(B) = \\{x \in X | f(x) \in B\\}$$ Prove the following claims: There are several claims that I am required to prove, but I am stuck on the first one, which is: $$B_1 \subseteq B_2 \Rightarrow f^{-1}(B_1) \subseteq f^{-1}(B_2)$$ So far I have done: $b \in (B_1 \subseteq B_2) \rightarrow b \in (f^{-1}(B_1) \subseteq f^{-1}(B_2))$ $\forall b [b \in B_1 \rightarrow b \in B_2]$ (Definition of a subset from my class' lecture notes.) $\rightarrow f^{-1}(b) \subseteq f^{-1}(b)$ I feel like this is the right direction, but that I must have missed a step because it feels too "easy". Any advice?

Think of the function $f$ and the relation $\subseteq$ as 'arrows'. Then you just have to start at the correct place and 'follow the arrows' correctly:

Let $x$ be in $f^{-1}(B_1)$. Then by definition, $f(x)$ is in $B_1$. Since $B_1$ $\subseteq$ $B_2$, $f(x)$ is in $B_2$. Then again by definition, $x$ is in $f^{-1}(B_2)$. QED.

_@Austin C : Drat, beat me._