Is the image of the intersection of preimage distributive and equal to the intersection of original set in the image? If $x\in\alpha^{-1}(A)\cap\alpha^{-1}(B)$, then we apply $\alpha$ to both sides and get $\alpha(x)\in\alpha(\alpha^{-1}(A)\cap\alpha^{-1}(B))$. Ok now this is the part where I am a bit unsure of. Can we just distribute $\alpha$ over the intersection and get $\alpha(x)\in A\cap B$?
Because it is true that $f(A)\cap f(B)\subseteq f(A\cap B)$. The intersection of two image is just a subset of the image of intersection so they are not equal, so that means $\alpha(\alpha^{-1}(A))\cap\alpha(\alpha^{-1}(B))\subseteq\alpha(\alpha^{-1}(A)\cap\alpha^{-1}(B))$ right? So if I distribute the $\alpha$ over the intersection, wouldn't I be getting a smaller set?
Edit: Ok the above is not correct. It is the case that $f(A\cap B)\subseteq f(A)\cap f(B)$ if the function is not injective and equal if injective.
 A: The comment at your edit is true.
Suppose that $f(x) = x^{2}$, $A = [-1,0]$ and $B = [0,1]$.
Then we have that $f(A\cap B) = \{0\}$ and $f(A)\cap f(B) = [0,1]$.
Consequently, $f(A)\cap f(B)\not\subseteq f(A\cap B)$.
On the other hand, if $f$ is injective, then we deduce that
\begin{align*}
y\in f(A)\cap f(B) & \Rightarrow (y\in f(A))\wedge(y\in f(B))\\\\
& \Rightarrow (\exists a\in A)(y = f(a))\wedge(\exists b\in B)(y = f(b))\\\\
& \Rightarrow y = f(a) = f(b)\\\\
& \Rightarrow a = b\in A\cap B\\\\
& \Rightarrow y\in f(A\cap B)
\end{align*}
and we are done.
Hopefully this helps.
A: If $x \in \alpha^{-1}[A] \cap \alpha^{-1}[B]$ then $\alpha(x) \in A$ from being in the first set and also $\alpha(x) \in B$ from being in the second set and so indeed $\alpha(x) \in A \cap B$. This is direct from the definitions.
This is also immediate if you know that inverse images distribute over intersections and unions so that
$$\alpha^{-1}[A] \cap \alpha^{-1}[B] = \alpha^{-1}[A \cap B]$$
This does not hold for forward images of a function in general, but you don't even need that for the conclusion that $\alpha(x) \in A \cap B$.
