Any function with the same polymorphic type as fmap must be equal to fmap?

Question

I'm reading the second edition of Programming in Haskell and I've came across this sentence:

... there is only one way to make any given parameterised type into a functor, and hence any function with the same polymorphic type as fmap must be equal to fmap.

This doesn't seem right to me, though. I can see that there is only one valid definition of fmap for each Functor type, but surely I could define any number of functions with the type (a -> b) -> f a -> f b which aren't equivalent to each other?

Why is this the case? Or, is it just a mistake by the author?

Answer 1

You've misread what the author was saying.

...any function with the same polymorphic type as fmap...

This means, any function with the signature

Functor f => (a -> b) -> f a -> f b

must be equivalant to fmap. (Unless you permit bottom values, of course.)

That statement is true; it can be seen quite easily if you try to define such a function: because you know nothing about f except that it's a functor, the only way to obtain a non-⊥ f b value is by fmapping over the f a one.

What's a bit less clear cut is the logical implication in the quote:

there is only one way to make any given parameterised type into a functor, and hence any function with the same polymorphic type as fmap must be equal to fmap.

I think what the author means there is, because a Functor f => (a -> b) -> f a -> f b function must necessarily invoke fmap, and because fmap is always the only valid functor-mapping for a parameterised type, any Functor f => (a -> b) -> f a -> f b will indeed also in practice obey the functor laws, i.e. it will be the fmap.

I agree that the “hence” is a bit badly phrased, but in principle the quote is correct.