Taking $n-1$ bits from hash function will also be hash function?

Question

If $\mathcal{H}=\left(\text{Gen},H\right)$ is a collision resistant hash family does $\mathcal{W}=\left(\text{Gen},W\right)$ where $W_{s}\left(x\right)$ denote the $n-1$ left most bits of $H_{s}\left(x\right)$ is also collision resistant hash family?

Im thinking that $\mathcal{H}$ would not be collision resistant hash family but I cant find an exact counter example of $x \neq x'$ such that $W_{s}\left(x\right) = W_{s}\left(x'\right)$. My guess was to prove using induction that we can deduce $k$ bits and still consided to be a collision resistant hash function, even if $k=n-1$ which will lead us to a function with image of 1 bit and that obviously wont be valid hash function.

Answer 1

If $\mathcal{H}=\left(\text{Gen},H\right)$ is a collision resistant hash family does $\mathcal{W}=\left(\text{Gen},W\right)$ where $W_{s}\left(x\right)$ denote the $n-1$ left most bits of $H_{s}\left(x\right)$ is also collision resistant hash family?

In general, no.

Let us assume that $\mathcal{H}'$ is a collision resistant hash family, and define $\mathcal{H}_k(x || b) = \mathcal{H}'_k(x) || b$, where $b$ is a single bit.

$\mathcal{H}$ is collision resistant (as a collision in $\mathcal{H}_k$ would imply a collision in $\mathcal{H}'_k$); however $\mathcal{W}$ is not, as $\mathcal{W}_k(00) = \mathcal{W}_k(01)$