# Category example question (Hungerford)

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?

• 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 $hf=g$. – Berci Feb 19 '14 at 23:26

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 $\langle f,h,g\rangle:f\to g$...
• 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 $g=p$? – Frudrururu Feb 27 '14 at 2:22