When I click on a link, how can I change the status of another component?

Question

When I click on a button, it redirects me to another component (#2) where there are two states and according to which state it shows one content or another. The default state of the component(#2) is true and what I need to know how to do is that clicking on the button of the component(#1) changes the state to false.

This is for a ReactJs app, im using function components, hooks, no class components

Thats the button code:
<Link to="/register">
<button className="btn btn-success btnVerde">
</button>
</Link>

Thats the component#2:
const [state, setState] = React.useState(true);
{state && <div1......../>}
{!state && <div2......./>}

I expect to click on the button in component#1, it redirecto to component#2 how default state is true, and change it to false

Answer 1

Thx to @rcoro that is how I solve the problem component#1:

const change = () => props.changingState(false);
<Link to="/register"><button onClick={change}className="btnPink">
         {t("talents_btnReserve")}
        </button>
        </Link>

component#2:

const showTalentRegister = () => setState(true);
const showCompanyRegister = () => setState(false);

useEffect(() => {
if(props.booleanState){
  showCompanyRegister()
}
else {
  showTalentRegister()
}

}, []);

Answer 2

you could handle the state in the component1, and in your onClick call setState and update the value with the oposite, eg:

<button onClick={() => setState({state: !state})}>Click</button>

and receive the value as a prop in component2, eg:

const Component2 = ({showCustom}) => {
  if(showCustom) {
    return <span> Custom </span>;
  }
  return <span>No Custom</span>;
};