Categories for the Working Mathematician, S. Mac Lane, 3.2, ex. 1:

Let functors $K, K'\colon D\to \mathbf{Set}$ have representations $\left<r,\psi\right>$ and $\left<r', \psi'\right>$, respectively. Prove that to each natural transformation $\tau\colon K\stackrel{*}{\rightarrow} K'$, there is a unique morphism $h\colon r' \to r$ of $D$ such that $$ \tau\circ\psi = \psi' \circ D\,(h, -)\colon D\,(r,-)\stackrel{*}\longrightarrow K' $$

Here's my attempt:

Since $\psi'$ is a representation, it's an isomorphism, hence there exists an inverse natural transformation $\psi'^{-1}$, obtained by reversing the arrows. Consider composition $$ \psi^{\prime-1}\circ\tau\circ\psi\colon D\,(r,-)\to D\,(r',-) $$ It's a natural transformation between covariant hom-functors. It follows from Yoneda lemma there exists unique $h\colon r'\to r$ such that $\psi^{\prime-1}\circ\tau\circ\psi = D\,(h,-)$. Composing with $\psi'$ on the left yields the thesis.

Now, what seems disturbing is that the above argument does not rely on the fact $\psi$ is isomorphic. Am I being paranoid? Or have I missed something?


1 Answer 1


Your proof is correct. It is an immediate consequence of the Yoneda Lemma. One only needs that $\psi : D(r,-) \to K$ is some natural transformation.


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