The Unapologetic Mathematician

The Weak Yoneda Lemma

The Yoneda Lemma is so intimately tied in with such fundamental concepts as representability, universality, limits, and so on, that it’s only natural for us to want to enrich it. Unfortunately, we’re only ready to talk about bijections of sets, not about isomorphisms of $\mathcal{V}$-objects. So this will give back Yoneda when we consider $\mathbf{Set}$-categories, but in general it won’t yet have the right feel.

So let’s say we’ve got a $\mathcal{V}$-functor $F:\mathcal{C}\rightarrow\mathcal{V}$, an object $K\in\mathcal{C}$, and a natural transformation $\eta:\hom_\mathcal{C}(K,\underline{\hphantom{X}})\rightarrow F$. We can construct the composite $\mathbf{1}\rightarrow\hom_\mathcal{C}(K,K)\rightarrow F(K)$, giving an element of the underlying set of $F(K)$. The weak Yoneda Lemma states that this construction gives a bijection between $\mathcal{V}\mathrm{-nat}(\hom_\mathcal{C}(K,\underline{\hphantom{X}}),F)$ — the set of $\mathcal{V}$-natural transformations from the $\mathcal{V}$-functor represented by $K$ and the $\mathcal{V}$-functor $F$ — and the underlying set of the $\mathcal{V}$-object $F(K)$.

We have the function going one way. We must now take an “element” $\xi:\mathbf{1}\rightarrow F(K)$ and build from it a natural transformation with components $\eta_C:\hom_\mathcal{C}(K,C)\rightarrow F(C)$. And we must also show that it inverts the previous function.

First off, since $F$ is a functor we have an arrow $\hom_\mathcal{C}(K,C)\rightarrow\hom_\mathcal{V}(F(K),F(C))$, which is the same as $F(C)^{F(K)}$. Now we can use the arrow $\xi$ to get an arrow $F(C)^\mathbf{1}$, which is isomorphic to $F(C)$. Every step here is natural in each variable by the litany of natural maps we laid down.

Now, if we compose this natural isomorphism with the identity arrow, it’s not hard to see that we get back $\xi$. In fact, the identity arrow $i_K:\mathbf{1}\rightarrow\hom_\mathcal{C}(K,K)$ followed by the application of $F$ gives the identity arrow $i_{F(K)}:\hom_\mathcal{V}(F(K),F(K))$. But then the exponential $\hom_\mathcal{V}(\xi,1_{F(K)})$ just says to compose $\xi$ with the identity on $F(K)$, and we’re left with $\xi$.

For the other direction — that starting with an isomorphism, constructing an element, and then constructing another isomorphism gives us back the isomorphism we started with — I refer you to this diagram:

We start with the isomorphism $\eta$ and construct the isomorphism along the lower-left of the diagram. The top row of the diagram is the identity (show it), and so the upper-right of the diagram is the original isomorphism $\eta$. I leave it to you to show that each of the three squares commute, and this our two constructions invert each other.

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Posted by John Armstrong | Category theory

1 Comment	»

1. [...] Strong Yoneda Lemma We gave a weak, “half-enriched” version of the Yoneda Lemma earlier. Now it’s time to pump it up to a fully-enriched [...]

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