4
$\begingroup$

Let $F$ be a free object on the set $X$ (with $i:X\rightarrow F$) in a concrete category $\mathcal{C}$. Define a new category $\mathcal{D}$ as follows. The objects of $\mathcal{D}$ are all maps of sets $f:X\rightarrow A$, where $A$ is (the underlying set of) an object of $\mathcal{C}$. A morphism in $\mathcal{D}$ from $f:X\rightarrow A$ to $g:X\rightarrow B$ is defined to be a morphism $H:A\rightarrow B$ of $\mathcal{C}$ such that the diagram commutes, i.e. $hf=g$.

This is just an example of some category in Hungerford algebra textbook. I am trying right now to show as an exercise for myself that $\mathcal{D}$ is a category. I see it has set of objects $$\mathrm{Ob}(\mathcal{D})=\{ f:X\rightarrow A \mid A \in \mathrm{Ob}(\mathcal{C}),\, f \text{ is a set map}\}$$ and if we have $f,g \in \mathrm{Ob}(\mathcal{D})$ with $f:X\rightarrow A$ and $g:X\rightarrow B$ then $$\mathrm{Hom}(f,g)=\{ h:A\rightarrow B \mid h\in \mathrm{Hom}_{\mathcal{C}}(A,B) \text{ where } hf=g\}.$$ (Abusing set notation at least for $\mathrm{Ob}(\mathcal{D})$.)

We require $\mathrm{Hom}(f,g)\cap \mathrm{Hom}(h,k)=\varnothing$ if $(f,g)\neq(h,k)$.

Say $\phi \in \mathrm{Hom}(f,g)\cap \mathrm{Hom}(h,k)$ where $$f:X\rightarrow A,\, g:X\rightarrow B,\, h:X\rightarrow C,\, k:X\rightarrow D.$$ Then by definition of $\mathrm{Hom}(f,g), \mathrm{Hom}(h,k)$, $\phi \in \mathrm{Hom}_{\mathcal{C}}(A,B)\cap \mathrm{Hom}_{\mathcal{C}}(C,D)$ so we have a contradiction unless $A=C$ and $B=D$. My goal now is to show that $f=h$ and $g=k$ unless I am misunderstanding something along the way. Here is where I am trying to figure out how this follows. So I have commutative diagrams $\phi f=g$, $\phi h=k$. Then as $F$ is free on $X$, I have unique maps $\overline{f},\overline{h}\in \mathrm{Hom}_{\mathcal{C}}(F,A)$ where $\overline{f}i=f$ and $\overline{h}i=h$ and similarly there are unique maps $\overline{g},\overline{k}\in \mathrm{Hom}_{\mathcal{C}}(F,B)$ such that $\overline{g}i=g$ and $\overline{k}i=k$. How does one show that the sets must be disjoint?

$\endgroup$
  • 4
    $\begingroup$ If homsets cannot intersect, the definition of such a morphism should be in a precise way a triple $\langle f,h,g\rangle\in \hom(f,g)$ such that $hf=g$. $\endgroup$ – Berci Feb 19 '14 at 23:26
2
$\begingroup$

Well, in a sense, formally, you are absolutely right.

However, it is not meant so strictly. Either it is only a vague usage of symbols, or the exact definition of category in the context didn't actually require the disjointness of distinct homsets.

If we require that, we have to define morphisms of $\mathcal D$ as triples $\langle f,h,g\rangle\ \in \hom(f,g)$, of course, which satisfy $hf=g$. For shorthand, as domain and codomain are usually always clear from context, we are allowed to write $h:f\to g$ instead of $\langle f,h,g\rangle:f\to g$...

$\endgroup$
  • $\begingroup$ So it can kind of be seen like a cartesian product so if $\left< f,h,g\right>=\left<k,m,p\right>$ then $f=k,h=m,$ and $g=p$? $\endgroup$ – Frudrururu Feb 27 '14 at 2:22

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.