My question relates to my understanding of Arrow's Impossibility Theorem and ranked choice. It seems to me that the requirements on social choice functions are too strict. A social choice function cannot result in ties. This sounds unreasonable to me.
Let two voters with preferences $a > b > c > d$ and $a > c > b > d$. It seems that any "reasonable" choice function would choose $a > b = c > d$. Any other output would be making an arbitrary distinction between $b$ and $c$. So my question is: if we relax the output of our social choice functions to allow for ties, is there a social choice function that satisfies Arrow's criteria? Is there proof that no such function exists?
As an addendum, I tried a "trivial" such function but it doesn't seem to work. Let $f$ a social choice function which if every voter prefers $c_1 > c_2$ then in the final output $c_1 > c_2$; otherwise $c_1 = c_2$.
This seems to fail by the transitivity of ties. So, for two preferences $a > b > c$ and $c > a > b$ then $a > b$ but $c = a$ and $c = b$ which is a contradiction. If we were to choose, $a > b = c$ we would be violating the independence of irrelevant alternatives. As, if we were to delete $b$ the result "should" be $a = c$. Is there a way to work around this?