Closed rokn closed 2 years ago
I'm not sure if this is correct, but it should be. Some lines before that it is said that f1 and f2 are of types a -> a' and b -> b' so it's only logical in the definition of bimap their corresponding types to be the same.
When I read this section I first thought that the example was wrong, too. After reading it again closely it made sense.
I guess we can now close this, feel free to reopen if you feel like I was wrong.
I'm not sure if this is correct, but it should be. Some lines before that it is said that f1 and f2 are of types a -> a' and b -> b' so it's only logical in the definition of bimap their corresponding types to be the same.