I'm having trouble understanding how the existence of a MAC implies the existence of an OWF.
If the MAC protocol is $(\operatorname{Gen}, \operatorname{MAC}, \operatorname{Ver})$ such that for any adversary $\mathcal{A}$,
\begin{gather*} \underset{k \sim \operatorname{Gen}(1^n)}\Pr[\operatorname{Ver}(k, \operatorname{MAC}(k, m)) = m] = 1, \qquad\text{and} \\ \underset{k \sim \operatorname{Gen}(1^n)}\Pr[\mathcal{A}(1^n) = (m,t) \mathrel{\text{such that}} \operatorname{MAC}(k, m) = t] < \text{negligible}. \end{gather*}
I am imagining the OWF I would construct to look something like $f(x) = \operatorname{MAC}(x, 0)$. If an adversary can invert $f$, then given the 'tag' $t$ of the zero string they are likely to be able to find some key $k$ such that $\operatorname{MAC}(k, 0) = t$. They could then break the security property of the MAC by using $k$ to forge new messages.
The problem I am running into is: just because inverting $f$ gives some key $k$ where $\operatorname{MAC}(k,0) = t$, it isn't necessarily the same with the challenger used, and may just happen to have the same tag at 0. How can I get around this?