Let $X$, $Y$ be Banach spaces with the same cardinality and $f: X \rightarrow Y$ be a surjection. Is it possible for $\varepsilon >0$ to find a bijection $g: X \rightarrow Y$ such that 
 $\|f-g\|\_{sup} < \varepsilon$?



Find a bijection near to a given surjection
Let $\kappa$ be the common cardinality of $X$ and $Y$. Fix well-orderings of $X$ and $Y$ of order-type $\kappa$ (i.e., each element has strictly fewer than $\kappa$ predecessors). Note that every open ball in either space also has cardinality $\kappa$ (because the space is covered by the countably many dilations, by integer factors, of any open ball around 0). Now define the desired bijection $g$ by the following transfinite back-and-forth construction of length $\kappa$. 


At every second stage, you take the first (in your well-ordering) element $x$ of $X$ to which no $g$-image has yet been assigned and you assign to it some $g(x)$ that is within $\epsilon/2$ of $f(x)$ and is not yet in the image of $g$. You can do that because the number of values assigned for $g$ in previous steps is $<\kappa$ and the $\epsilon/2$-ball around $f(x)$ has cardinality $\kappa$.


At the other steps, you take the first $y\in Y$ that has not yet been put into the range of $g$ and you set $g(z)=y$ for some (say the first) $z\in X$ that has not been previously assigned a $g$-image and that has $f(z)$ within $\epsilon/2$ of $z$. Such a $z$ exists because the $\epsilon/2$-ball around $z$ has cardinality $\kappa$, $f$ is surjective, and fewer than $\kappa$ elements of $X$ have had $g$-values previously assigned.


Because of the heavy use of the axiom of choice, I can't claim any sort of good behavior for this $g$, like linearity, continuity, or even Borel-ness. Notice also that the result has very little to do with Banach spaces. It works for any metric spaces in which the balls have the same cardinality as the whole space.


