Suppose you have two objects $A$ and $A'$ in a category $\mathfrak{C}$, and morphisms $i_C:\text{Mor}(C,A)\to\text{Mor}(C,A')$ for any object $C\in\mathfrak{C}$. Show that the $i_C$ are induced from a unique morphism $g:A\to A'$. More precisely, show that there is a unique morphism $g:A\to A'$ such that for all $C\in\mathfrak{C}$, $i_C$ is $u\to g\circ u$.

That $g$ is unique seems wrong to me! Taking a simple example in $\text{Set}$ would show this. Any help would be greatly appreciated.

Edit: The exercise says "The morphisms $i_C:\text{Mor}(C,A)\to\text{Mor}(C,A')$ commute with the maps of the form $\text{Mor}(C,A)\to\text{Mor}(B,A)$." I don't know what this means.

  • $\begingroup$ Take $C=A$, and consider the identity $\operatorname{Id}:A\rightarrow A$. $\endgroup$ – Joe Johnson 126 Sep 30 '14 at 11:28
  • 2
    $\begingroup$ I'm curious about your example though, seems like you misunderstood some part from the exercise. $\endgroup$ – roman Sep 30 '14 at 11:30
  • $\begingroup$ @roman- Please check the edits. $\endgroup$ – algebraically_speaking Sep 30 '14 at 11:37
  • 1
    $\begingroup$ A map $f\colon B\to C$ induces a map $f_A^*\colon\operatorname{Mor}(C,A)\to\operatorname{Mor}(B,A)$ defined by $f_A^*(u)=u\circ f$. The statement in the edit is that $i_B\circ f_A^*=f_{A'}^*\circ i_C$ for any $f\colon B\to C$ - this property is crucial for the statement you want to prove. $\endgroup$ – mdp Sep 30 '14 at 13:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.