Assume that $F \in C^{\wedge}$ is represented by $X_0 \in C$. Then $\text{Hom}_{C^{\wedge}}(h_C(X_0, F)) \simeq F(X_0)$ gives an element $s_0 \in F(X_0)$. Moreover, for any $Y \in C$ and $t \in F(Y)$ there exists a unique morphism $f: X_0 \to Y$ suc that $t = F(f)(s_0)$.

I know that the Yoneda Lemma bijections are give by:

$$ \varphi : \text{Hom}_{C^{\wedge}}(h_C(X), F) \to F(X) \\ \varphi(f) = f_X(\text{id}_X) \\ \psi(s)_Y = F(\cdot)(s) $$

But there useage of it seems to be going the opposite direction. Please help me understand what they mean.


  • $\begingroup$ Aside: $\hat{C} \hat C \widehat{C} \widehat C$ gives $\hat{C} \hat C \widehat{C} \widehat C$ $\endgroup$ – user14972 Oct 19 '17 at 19:06
  • $\begingroup$ @Hurkyl not sure what you mean there. $\endgroup$ – StudySmarterNotHarder Oct 19 '17 at 19:31
  • $\begingroup$ Those are latex codes to achieve the usual typesetting for the presheaf category. $\endgroup$ – user14972 Oct 19 '17 at 20:04
  • $\begingroup$ @Hurkyl Actually, the book really uses the $C^{\wedge}$ notation for the category of presheaves, and $C^{\vee}$ for the category of (covariant) functors $C^{op}\to Set^{op}$. $\endgroup$ – Arnaud D. Oct 20 '17 at 8:29

$F$ being represented by $X_0$ means that there is a natural isomorphism $\mu:h_C(X_0)\Rightarrow F$. In particular, $\mu\in Hom_{C^\wedge}(h_C(X_0),F)$; then the $s_0$ here is simply $\varphi(\mu)$.

The second sentence then basically says that for all $Y$, the function $$(h_C(X_0))(Y)=Hom_C(Y,X_0)\to F(Y):f\mapsto F(f)(s_0)$$ is a bijection$^1$; but this function is actually the definition of $\psi(s_0)_Y$, and thus it simply follows from the fact that $\psi(s_0)=\psi(\varphi(\mu))=\mu$, because $\varphi$ and $\psi$ are inverse bijections.

$1$: I don't know if that's the source of your confusion, but there seems to be a mistake in the book: here the $f:X_0\to Y$ should really be $f:Y\to X_0$, otherwise it wouldn't make sense to compute $F(f)(s_0)$ since $F$ is assumed to be contravariant.

  • $\begingroup$ Could you use the notation $F(f)(s)$ in there? $\endgroup$ – StudySmarterNotHarder Oct 19 '17 at 19:34
  • $\begingroup$ I'm not following how $\psi(s_0)_Y$ is a bijection $\endgroup$ – StudySmarterNotHarder Oct 19 '17 at 20:13

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