TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Representing objects are unique up to a unique isomorphism compatible with their universal elements

Statement

Let $F : C \to S e t$ have universal elements $(R, u)$ and $(R^{'}, u^{'})$ . There is a unique isomorphism $i : R \to R^{'}$ satisfying $F (i) (u) = u^{'}$ .

If instead $P : C^{o p} \to S e t$ has universal elements $(R, u)$ and $(R^{'}, u^{'})$ , there is a unique isomorphism $i : R \to R^{'}$ satisfying $P (i) (u^{'}) = u$ .

Hence a representing object is unique up to the unique isomorphism compatible with the chosen universal elements, in either variance.

Facts & Assumptions

Given: A locally small category $C$ and the two universal elements for the same covariant functor or presheaf appearing in the statement.

[L1]

For a covariant universal element $(R, u)$ , every $x \in F (c)$ has a unique expression $F (f) (u)$ with $f : R \to c$ ; for a presheaf universal element, every $x \in P (c)$ has a unique expression $P (f) (u)$ with $f : c \to R$ (A representation is equivalently a universal element with a unique factorisation property).

[F1]

A morphism is an isomorphism when it admits a two-sided inverse (Isomorphism, groupoid, and connected category).

Proof

technique · direct

1.1

In the covariant case, apply [L1] for $(R, u)$ to $u^{'} \in F (R^{'})$ and for $(R^{'}, u^{'})$ to $u \in F (R)$ , obtaining unique morphisms $i : R \to R^{'}$ and $j : R^{'} \to R$ with $F (i) (u) = u^{'}$ and $F (j) (u^{'}) = u$ .

givenL1

1.2

In the presheaf case, [L1] applied to $u \in P (R)$ and $u^{'} \in P (R^{'})$ gives unique $i : R \to R^{'}$ and $j : R^{'} \to R$ with $P (i) (u^{'}) = u$ and $P (j) (u) = u^{'}$ .

givenL1

2.1

Functoriality gives $F (j \circ i) (u) = u = F (1_{R}) (u)$ ; uniqueness in [L1] for $(R, u)$ gives $j \circ i = 1_{R}$ . Similarly, $i \circ j = 1_{R^{'}}$ .

step 1.1L1

3.1

By [F1], $i$ is an isomorphism. Any compatible morphism $k : R \to R^{'}$ satisfies $F (k) (u) = u^{'}$ and therefore equals $i$ by [L1], proving uniqueness even among all compatible morphisms.

step 1.1step 2.1L1F1

4.1

Contravariant functoriality and the same uniqueness argument give $j \circ i = 1_{R}$ and $i \circ j = 1_{R^{'}}$ ; [F1] and uniqueness in [L1] then make $i$ the unique compatible isomorphism.

step 1.2L1F1∎

Depends on

Used by

Dependency tree · two levels

7 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Emily Riehl, Category Theory in Context, Corollary 2.3.2 (standard reference, not scraped)
Justin Campbell, Harvard Math 55b tutorial notes, Corollary 1.2.1 (standard reference, not scraped)