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copied!<p>Not sure I understand what was the question but yes, you are right, monad in Haskell is defined as a triple:</p> <pre><code>m :: * -> * -- this is endofunctor from haskell types to haskell types! return :: a -> m a (>>=) :: m a -> (a -> m b) -> m b </code></pre> <p>but common definition from category theory is another triple:</p> <pre><code>m :: * -> * return :: a -> m a join :: m (m a) -> m a </code></pre> <p>It is slightly confusing but it's not so hard to show that these two definitions are equal. To do that we need to define <code>join</code> in terms of <code>(>>=)</code> (and vice versa). First step:</p> <pre><code>join :: m (m a) -> m a join x = ? </code></pre> <p>This gives us <code>x :: m (m a)</code>.<br> All we can do with something that have type <code>m _</code> is to aply (>>=) to it:</p> <pre><code>(x >>=) :: (m a -> m b) -> m b </code></pre> <p>Now we need something as a second argument for (>>=), and also, from the type of join we have constraint <code>(x >>= y) :: ma</code>.<br> So <code>y</code> here will have type <code>y :: ma -> ma</code> and <code>id :: a -> a</code> fits it very well:</p> <pre><code>join x = x >>= id </code></pre> <h2>The other way</h2> <pre><code>(>>=) :: ma -> (a -> mb) -> m b (>>=) x y = ? </code></pre> <p>Where <code>x :: m a</code> and <code>y :: a -> m b</code>. To get <code>m b</code> from <code>x</code> and <code>y</code> we need something of type <code>a</code>.<br> Unfortunately, we can't extract <code>a</code> from <code>m a</code>. But we can substitute it for something else (remember, monad is a functor also):</p> <pre><code>fmap :: (a -> b) -> m a -> m b fmap y x :: m (m b) </code></pre> <p>And it's perfectly fits as argument for <code>join</code>: <code>(>>=) x y = join (fmap y x)</code>.</p>

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