Let $f$ be a bijective mapping from a set $A$ to a set $B$, and let $C \subsetneqq A$. Then the image of the restriction $f{\restriction_C}(C)$ is a proper subset of the codomain $B$. Proof: By contradiction: obviously is $f{\restriction_C}(C) \subset B$; suppose $f{\restriction_C}(C) = B$:
picked any $y$ from $B$ we get an $x \in C$ such that $y = f{\restriction_C}(x)$. By the definition of restriction of a map, we know that $f{\restriction_C} \subset f$, so $y = f(x)$;
$f$ is bijective so, for every $y \in B$ there exists $x \in A$ such that $y = f(x)$.
I don't know how to continue the proof. Intuition suggests me that these two condition lead to a contradiction, because I can choose $y \in B$ such that $y = f(x_1) = f(x_2)$, where $x_1 \in A \setminus C$ and $x_2 \in C$, and so $x_1 \neq x_2$, against the hypothesis that $f$ is injective. But how do I chose that $y$?
I would think of it this way.
Since $C\subsetneq A$, there is an $a\in A\setminus C$. Let $b:=f(a)$. If, on contrary, we had $f(C)=B$, then there would have to be a point $c\in C$ such that $f(c)=b$. But then $f(a)=f(c)$ and $a$ and $c$ are distinct. This contradicts that $f$ is a bijection.