Proposition 8.3.1.1 (03M3): Yoneda's Lemma, Strong Form

Proposition 8.3.1.1 (Yoneda's Lemma, Strong Form). Let $\operatorname{\mathcal{C}}$ be a category containing an object $X$. For every functor $\mathscr {F}: \operatorname{\mathcal{C}}\rightarrow \operatorname{Set}$, evaluation on the identity morphism $\operatorname{id}_{X} \in h^{X}(X)$ induces a bijection

\[ \operatorname{Hom}_{ \operatorname{Fun}( \operatorname{\mathcal{C}}, \operatorname{Set}) }( h^{X}, \mathscr {F} ) \rightarrow \mathscr {F}(X). \]

Proof. Fix an element $x \in \mathscr {F}(X)$. We wish to show that there is a unique natural transformation $\alpha : h^{X} \rightarrow \mathscr {F}$ which carries $\operatorname{id}_{X} \in h^{X}(X)$ to the element $x \in \mathscr {F}(X)$.

For any object $Y \in \operatorname{\mathcal{C}}$, every element $f \in h^{X}(Y) \in \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)$ can be obtained by evaluating the function $h^{X}(f): h^{X}(X) \rightarrow h^{X}(Y)$ on the object $\operatorname{id}_{X}$. It follows that, if $\alpha : h^{X} \rightarrow \mathscr {F}$ is a natural transformation satisfying $\alpha _{X}( \operatorname{id}_ X ) = x$, then it must satisfy the identity

\[ \alpha _{Y}(f) = \alpha _{Y}( h^{X}(f)(\operatorname{id}_ X) ) = \mathscr {F}(f)( h_ X( \operatorname{id}_ X ) ) = \mathscr {F}(f)(x). \]

This proves uniqueness. To establish existence, it will suffice to show that the collection of functions

\[ \alpha _{Y}: \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y) \rightarrow \mathscr {F}(Y) \quad \quad f \mapsto \mathscr {F}(f)(x) \]

determine a natural transformation from $h^{X}$ to $\mathscr {F}$. In other words, we must show that for each morphism $g: Y \rightarrow Z$ in $\operatorname{\mathcal{C}}$, the diagram of sets

\[ \xymatrix@C =50pt@R=50pt{ \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y) \ar [d]^{ \alpha _{Y} } \ar [r]^-{ g \circ } & \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Z) \ar [d]^{ \alpha _ Z} \\ \mathscr {F}(Y) \ar [r]^-{ \mathscr {F}(g) } & \mathscr {F}(Z) } \]

is commutative. This follows from the observation that, for every morphism $f: X \rightarrow Y$ of $\operatorname{\mathcal{C}}$, we have an equality $\mathscr {F}( g \circ f)(x) = (\mathscr {F}(g) \circ \mathscr {F}(f))(x)$ in the set $\mathscr {F}(Z)$. $\square$