No signature when using union type of functions

[OliverJAsh]

v2.1.4

{
    type Fn1 = (x: number) => number;
    type Fn2 = (x: any) => string;

    const fn = 1 as any as Fn1 | Fn2

    fn(1) // error: Cannot invoke an expression whose type lacks a call signature.
}

However, if I change Fn2 to type Fn2 = (x: number) => string; (thus removing the any), the error disappears.

I would like to understand why TypeScript can't handle the former case.

For context, I am trying to build an Option type, where Option = Some | None:

{
    class Some<T> {
        constructor (public t: T) {}
        map<T2>(f: (t: T) => T2): Some<T2> { return new Some(f(this.t)) }
    }

    class None {
        constructor () {}
        map(f: (t: any) => any): None { return new None() }
        get() { throw new Error('foo') };
    }

    type Option<T> = Some<T> | None;

    const option = new Some(1) as Option<number>;

    option.map(x => 1) // error: Cannot invoke an expression whose type lacks a call signature.
}

[mhegazy]

A union type has only gets signatures that are identical in all constituents. the reason is there is no meaningful way the compiler can combine these signatures, for the general case, and still grantee type safety. see #10620 for more details.