1-1 map on the $\{0,1\}^k$

Question

Let integer $k>0$ and let $\{0,1\}^k$ denote the set of all $1\times k$-dim vectors whose every coordinate is eithor 0 or 1, for example, $(0,1,1,0,\dots,1,0,0,1)$. For any such vector $\alpha$, let $\alpha_i$ denote it's $i$'s coordinate.

Let integer $k_0<k$ (or one can consider $\sqrt{k}<k_0\ll k$) and let $A\subseteq\{0,1\}^k$ denote the set of all $1\times k$-dim vectors whose every coordinate is eithor $0$ or $1$ and for every vector in $A$, the vector has value $0$ on more than $k/2+k_0$ many coordinates. Let $B\subseteq\{0,1\}^k$ denote the set of all $1\times k$-dim vectors whose every coordinate is eithor $0$ or $1$ and for every vector in $A$, the vector has value $1$ on more than $k/2+k_0$ many coordinates. Clearly $|A|=|B|,A\cap B=\emptyset$.

For every $x\in A$, is there exist an 1-1 map $f_x:A\rightarrow B$ such that for every $a\in A$, let $I_x^a\in [k]$ denote the set of coordinates where $a$ and $f_x(a)$ are different, the following holds:?

For every $i\in [I_x^a]$, $a_i=x_i$ and $f_x(a)_i\neq x_i$.

Answer 1

I assume words “more” and “less” to be non-strict, otherwise shift $k_0$ by 1.

Assume that the weight (=number of 1s) of $x$ is $k/2-k_0$. There are plenty of $a$s which agree with $x$ only at $2k_0$ positions (both have zeroes at those positions). Then $f_x(a)$ should differ from $a$ at all those positions in order to come to $B$. But this means that $f_x(a)=\mathbb 1-x$ for all those $a$, so the map is not bijective.