Is it possible to define $Hom(-, A): \mathcal C \to \mathbf {Set}$ as simply $Hom_{\mathcal C}(-, A)(\require{AMScd}\begin{CD}X @>f>> Y\end{CD}) = Hom_{\mathcal C^{\mathbf {op}}}(A,-)(\require{AMScd}\begin{CD}X @>{f^{\mathbf {op}}}>> Y\end{CD})$ ?

  • $\begingroup$ Sure. But this involves some small abuse of notation. $\endgroup$ – Zhen Lin Sep 7 '15 at 10:47
  • $\begingroup$ Isn't there a way to express the thought without redundantly rewriting all the mapping rules? $\endgroup$ – Michele De Pascalis Sep 7 '15 at 10:50
  • $\begingroup$ I don't really understand what your question is. Simply writing down the definitions (and I literally mean "By definition $g \circ f = f^{op} \circ^{op} g^{op}$, QED") gives the answer. What are you asking, precisely? $\endgroup$ – Najib Idrissi Sep 7 '15 at 11:08
  • $\begingroup$ The book I'm consulting (Abstract and Concrete Categories) takes the bother to redefine $Hom(-,A)$ explicitly after having both defined $Hom(A,-)$ and explained the Duality Principle, so I was wondering if there was some reason not to define it as I did. $\endgroup$ – Michele De Pascalis Sep 7 '15 at 11:12

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