Elaborate a bit more on Keith's answer.

**Puzzle 161**

1) **Preservation of composition:**

Suppose \\(h\in\mathcal{C}(c, c')\\) and \\((f,g)\\) is a morphism from \\((c, c')\\) to \\((d, d')\\) and \\((l, j)\\) is a morphism from \\((d, d')\\) to \\((e, e')\\), we have

\[
\begin{array}{ccc}
\mathrm{hom}\big((l, j)\circ(f, g)\big) h &=&\mathrm{hom}\big((l\circ_{op} f, j\circ g)\big)h\\\\
&:=&(j\circ g)\circ h \circ (l\circ_{op} f)\\\\
&=&(j\circ g)\circ h \circ (f\circ l)\\\\
&=&j\circ (g\circ h \circ f)\circ l\\\\
&:=&\mathrm{hom}\big((l, j)\big) (g\circ h \circ f)\\\\
&:=&\mathrm{hom}\big((l, j)\big) \circ \mathrm{hom}\big((f, g)\big) h\\\\
\end{array}
\]
This shows that \\(\mathrm{hom}\big((l, j)\circ(f, g)\big)=\mathrm{hom}\big((l, j)\big) \circ \mathrm{hom}\big((f, g)\big)\\).


2) **Preservation of identities:**

Suppose \\(h\in\mathcal{C}(c, c')\\) and \\( 1 _{c, c'}=(\mathrm{id}_c, \mathrm{id} _{c'})\\), then
\[
\begin{array}{ccc}
\mathrm{hom}(1_{c, c'}) h &:=&\mathrm{id} _{c'}\circ h\circ\mathrm{id} _{c}\\\\
&=&h
\end{array}
\]

Hence \\(\mathrm{hom}(1 _{c, c'})\\) is the identity map on the set \\(\mathcal{C}(c, c')\\), i.e. \\(\mathrm{hom}(1 _{c, c'})=1 _{\mathcal{C}(c, c')}\\).