Proof explanation: proving equivalent assertions and dealing with injectivity Let $E$ and $F$ be two sets and $f: E → F$. Prove that the following two assertions are equivalent:


*

*$f$ is injective.

*For all $A, B ⊆ E$, we have $f(A ∩ B) = f(A) ∩ f(B).$
So I am having trouble understanding the proof that assumes 2 and proves 1. I have provided below what the proof looks like, but I am confused why we are able to treat the function $f$ as a linear function because it is not known that the function $f$ is linear. I surrounded the part below that I am confused about with "**". 

Proof:
Assume $f(A ∩ B) = f(A) ∩ f(B)$ for all $A, B ⊆ E$. Let $x \neq y$ ∈ E. Then $\{x\} ∩ \{y\} = ∅$. So:
$**{ f(x)} ∩ { f(y)} = f({x}) ∩ f({y}) = f({x} ∩ {y}) = f(∅) = ∅**$

We begin with a quick observation. Let A, B ⊆ E. Let x ∈ A ∩ B. Then x ∈ A and so
$f(x) ∈ f(A)$. Similarly $f(x) ∈ B$ so $f(x) ∈ A ∩ B$. Therefore, $f(A ∩ B) ⊆ f(A) ∩ f(B)$ for any $f$ .
Assume first that $f$ is injective. Let $A, B ⊆ E$. Let $x ∈ f(A) ∩ f(B)$. There exists $z ∈ A$ and $y ∈ B$ such that $x = f(z) = f(y)$. Since $f$ is injective, $y = z$. Thus $y ∈ A ∩ B$. So $x = f(y) ∈ f(A ∩ B)$. This proves that
$f(A ∩ B) = f(A) ∩ f(B).$
 A: You're making use of the assumption in the bolded line. In terms of the set notation you provided, $A = \{x\}$ and $B = \{y\}$. So by the assumption,
\begin{align*}
f(A) \cap f(B) = f(x) \cap f(y) = f(\{x\} \cap \{y\}) = f(\emptyset)= \emptyset
\end{align*}
Spelled out, the first equality follows from our definition of $A$ and $B$. The second equality follows from the assumption. The third equality is true since $x\neq y$. The last equality follows from the definition of an image
\begin{align*}
f(\emptyset) = \{f(t) : t \in \emptyset\} = \emptyset
\end{align*}
A: $f$ is not assumed to be linear.  $f(A)$ where $A$ is a set is not a single input output value.  $f(A)$ is a set $= \{ f(x) | x \in A\}$.  
So $f(\{x\}) = \{f(x)\}$ .  $f(\{y\}) = \{f(y)\}$.  
$f(y) \ne f(x)$ (because $f$ is injective) so $f(\{x\}) \cap f(\{y\}) = \{f(x)\} \cap \{f(y)\} = \emptyset$.
$f(\emptyset) = \emptyset$ because $f(\emptyset) = \{f(x)|x \in \emptyset\} = \emptyset$.
$x \ne y$ so $\{x\} \cap \{y\} = \emptyset$.
So $\{f(x)\}\cap \{f(y)\} = f(\{x\}) \cap f(\{y\} ) = \emptyset$
And $f(\{x\} \cap \{y\})= f(\emptyset) = \emptyset$.
So $\{f(x)\}\cap \{f(y)\}= f(\{x\} \cap \{y\})$
has nothing to do with $f$ being or not being linear.
In general $f(A \cap B) = \{f(x)| x \in A \cap B\} = \{f(x)| x \in A\} \cap \{f(x)| x \in B\} = f(A) \cap f(B)$.
Nothing to do with linearity.
A: Writing something like $f(x)\cap f(y)$ is a (gross) abuse of language and can be very confusing.
I find it convenient to emphasize that $f(A)$ is not the at the same level as $f(x)$ and often use a special notation:
$$
f^{\to}(A)=\{f(a):a\in A\}
$$
Note that $f^{\to}$ is a map from $P(E)$ to $P(F)$ (power sets).
A general property of $f^{\to}$ is that
$$
f^{\to}(A\cap B)\subseteq f^{\to}(A)\cap f^{\to}(B)
$$
Indeed, if $x\in A\cap B$, then $x\in A$ and $x\in B$, so, by definition, $f(x)\in f^{\to}(A)$ and $f(x)\in f^{\to}(B)$.
The converse inclusion does not generally hold. The statement you have to prove is that the converse inclusion holds for every pair of subsets $A$ and $B$ if and only if $f$ is injective.
Suppose $f$ is not injective. Then there are $x,y\in E$ with $x\ne y$ and $f(x)=f(y)=z\in F$. Let $A=\{x\}$ and $B=\{y\}$. Then


*

*$A\cap B=\emptyset$

*$f^{\to}(A)=\{f(x)\}=\{z\}$

*$f^{\to}(B)=\{f(y)\}=\{z\}$


Thus $f^{\to}(A\cap B)=f^{\to}(\emptyset)=\emptyset$, whereas $f^{\to}(A)\cap f^{\to}(B)=\{z\}\ne\emptyset$. Hence, when $f$ is not injective, the equality $f^{\to}(A\cap B)\subseteq f^{\to}(A)\cap f^{\to}(B)$ does not hold for every $A,B\in P(E)$.
Now suppose $f$ is injective and take any $A,B\in P(E)$. We have to prove that $f^{\to}(A\cap B)\supseteq f^{\to}(A)\cap f^{\to}(B)$ (because the reverse inclusion always holds).
Let $z\in f^{\to}(A)\cap f^{\to}(B)$. Then there are $x\in A$ and $y\in B$ with $z=f(x)$ (that is, $z\in f^{\to}(A)$) and $z=f(y)$ (that is, $z\in f^{\to}(B)$). By injectivity of $f$, we conclude $x=y\in A\cap B$, so $z\in f^{\to}(A\cap B)$. Hence, when $f$ is injective, the equality $f^{\to}(A\cap B)\subseteq f^{\to}(A)\cap f^{\to}(B)$ holds for every $A,B\in P(E)$.

The above is an extended version of the proof you have, but I believe this point of view will clarify it.
