The injective image of complements is the complement of images

up:: The image of the complement contains the complement of the images

Note that, by the definition of an Injective Function, we have that¹

$$\exists x\in X\setminus B\mid f(x)=y⟺(\exists x\in X\mid f(x)=y)\wedge (\nexists x\in B\mid f(x)=y)$$

thus proving the reciprocal $f(X)\setminus f(B)\subseteq f(X\setminus B)$.

Footnotes

1. That is due to $\forall y\in Y,\exists !x\in X\mid f(x)=y$. ↩