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')}\$$.