Weighted limits

Question

I have a very trivial question about the page 80 here: how this shape of $W$ $$W:2\to \mathbf{Set}$$ with $$\ast\sqcup\ast\to \ast$$ implies that the components of $$W\Rightarrow\cal{M}(m,f)$$ are these two arrows $$h:m\to a,k:m\to a.$$

Perhaps that I've entirely forgotten what a natural transformation is for this case! I understand that the arrows should be 2 in number. But I do not understand the direction of $$h,k$$ as from $m\to a,$ though.

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Answer 1

Let's write down all of the definitions from the text first:

First, $ \mathcal M $ is any category, $ a $ and $ b $ are any objects in $ \mathcal M $, and $ f $ is any morphism from $ a $ to $ b $ in $ \mathcal M $. Next, $ \mathbf 2 $ (which is in blackboard bold in the original, but \mathbb doesn't work with 2 here, so I'm using bold) is a category with two objects and one nontrivial morphism, so a functor from $ \mathbf 2 $ is essentially a morphism in the target category. In particular, the text identifies $ f \colon a \to b $ in $ \mathcal M $ with a functor $ f \colon \mathbf 2 \to \mathcal M $. Meanwhile, $ W \colon \mathbf 2 \to \operatorname { Set } $ is identified with the unique morphism from a $ 2 $-element set $ \ast \sqcup \ast $ to a $ 1 $-element set $ \ast $. Since I think that $ \ast \sqcup \ast $ is not easy to use, I'll write this set as $ 2 $ (no bold or blackboard bold) and write $ \ast $ as $ 1 $ to go with it, so $ W \colon 2 \to 1 $. Finally, $ m $ is another object of $ \mathcal M $.

Now we need to understand what $ \mathcal M ( m , f ) $ means. If $ f $ were an object in $ \mathcal M $, then this would be the set of morphisms from $ m $ to $ f $ in $ \mathcal M $, but unfortunately, $ f $ is a morphism of $ \mathcal M $ instead. But since $ f \colon a \to b $, we can understand $ \mathcal M ( m , f ) $ as the function from $ \mathcal M ( m , a ) $ to $ \mathcal M ( m , b ) $ induced by composition with $ f $. Then we can identify this with a functor $ \mathcal M ( m , f ) \colon \mathbf 2 \to \operatorname { Set } $. This is good, because now it makes sense to talk about a natural transformation from $ W $ to $ \mathcal M ( m , f ) $.

So what is such a natural transformation? It consists of a morphism for each object of $ \mathbf 2 \, $ and a commutative square for each morphism of $ \mathbf 2 $, and the commutative square for the nontrivial morphism of $ \mathbf 2 $ has all of the information. The sides of this commutative square are the morphisms $ W \colon 2 \to 1 $ and $ \mathcal M ( m , f ) \colon \mathcal M ( m , a ) \to \mathcal M ( m , b ) $ in $ \operatorname { Set } $, so the top and bottom are morphisms $ x \colon 2 \to \mathcal M ( m , a ) $ and $ y \colon 1 \to \mathcal M ( m , b ) $. So $ x $ consists of two morphisms from $ m $ to $ a $ in $ \mathcal M $, which the paper calls $ h $ and $ k $. And $ y $ consists of one morphism from $ m $ to $ b $ in $ \mathcal M $, which is both $ f \circ h $ and $ f \circ k $ if and only if the square is commutative. So in the end, we have morphisms $ h , k \colon m \to a $ such that $ f \circ h = f \circ k $.