The Yoneda lemma and fully faithful functors
Recover natural transformations from identity arrows and track the variance of represented presheaves.
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The question for today
What you’ll understand
- Construct the Yoneda bijection for a contravariant set-valued functor.
- Prove the reconstruction formula satisfies naturality.
- Count natural transformations between represented presheaves using full faithfulness.
The Yoneda lemma says that a compatible rule on all arrows into an object is already determined by its value on that object's identity. Its strength comes from the compatibility assumption: it converts infinitely many possible component values into one datum. We use the presheaf convention consistently, so the represented functor has arrows into its representing object. A small group category and a three-point order will make both the reconstruction and its variance visible.
Evaluation at the identity loses no information
Assume is small and is a presheaf. Put . Yoneda gives a bijection between natural transformations and elements of . It sends a transformation to its component at evaluated on the identity arrow. The smallness assumption ensures that the functor category has set-sized Hom collections without further universe bookkeeping; the lemma itself also admits a locally small formulation.
Given , reconstruct the component at by sending to . The direction is correct because reverses the arrow. Naturality of any proposed transformation applied to the arrow forces exactly this formula. Hence no second transformation can have the same value on the identity, and every choice of that value produces a transformation once the naturality check is made.
Naturality in the probes and in the data
For an arrow , the presheaf law gives . Applying this identity to proves that the reconstruction commutes with precomposition of probe arrows. Evaluation and reconstruction are inverses because identities act trivially and because the original naturality square determines each component. This is a proof of the bijection, rather than an inference from a coincidental count in finite examples.
The bijection is also natural in and in . A natural transformation carries the chosen element to , matching composition with . An arrow induces postcomposition ; precomposing a transformation out of then corresponds to applying . These directions explain why changing the representing object can reverse arrows in a Hom expression.
The Yoneda embedding remembers every arrow
The assignment extends covariantly to a functor by postcomposition. Taking in Yoneda identifies natural transformations with arrows . Thus is full and faithful: it gives a bijection on every Hom set. Faithfulness detects equality of arrows, while fullness says every transformation between these represented presheaves comes from an original arrow.
Full faithfulness does not say that every presheaf is represented. It also does not force distinct but isomorphic source objects to become unequal in every possible presentation. Instead it recovers the entire arrow structure between the chosen represented objects. Using covariant representables would reverse the direction of the corresponding embedding. Writing the presheaf convention explicitly prevents a false answer even in a tiny ordered example.
Work it through
Four transformations from a regular three-cycle
Let have one object and arrows under addition. Let the presheaf have values , with adding modulo three to the numerical elements and fixing . Count the transformations from the representable presheaf to .
The representable has the three arrows as its elements; precomposition by adds . Because the group is abelian, the stated maps satisfy the contravariant composition law as well as the identity law.
Choose the value of a transformation on the identity residue zero. Reconstruction sends a residue to . For the three numerical choices of , this is one of the three translations of the cycle.
For , reconstruction is the constant map to the fixed element. All four maps satisfy equivariance with every group arrow, and Yoneda rules out any others. No map is a natural isomorphism, since its component has three inputs and four possible outputs.
There are exactly four natural transformations: three translated cycles and one constant map to the fixed point.
The living notebook
Explore the model
Interactive lab · modular
Reconstruct a cycle map from one value
Pause and Reset, then set modulus to 3 and additive step to 1. Two Steps select residue 2. On paper choose the image of identity to be 2 and predict the transformation's images of residues 0, 1, and 2: they are 2, 0, and 1. Compare these with the regular orbit. The fixed point * of the worked presheaf is not drawn, and the diagram does not enumerate general presheaves. Reset and set both controls before any fresh comparison; changing a control retains the clock.
- Residue r0
- 0
- Orbit period · m/gcd(m,a)
- 12
- Distinct residues visited
- 1 of 12
Addition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forOne selected value reconstructs an equivariant map, illustrating the identity-evaluation mechanism in Yoneda.
Pause & reason
Check your understanding
Your turn
Try it yourself
In the same chain category, compute directly from its components and compare the result with Yoneda.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
List the component sets at zero, one, and two. A function with empty domain is unique, including when its target is nonempty.
Compare with the solution
At zero and one, both presheaves have singleton values, so the component functions are unique. At two, is empty and is a singleton, so again there is one function. The naturality squares commute because each route is the only available function of its type. There is exactly one transformation, matching the unique arrow . This contrasts with the empty set of transformations in the reverse direction.
Keep the idea
Take it with you
- Yoneda reconstructs a transformation from its value on one identity arrow.
- For presheaves, arrows into the representing object are transported by precomposition.
- The Yoneda embedding is full and faithful, but not every presheaf is represented.
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