**Puzzle 161**

![homfunctor preservation rules](http://aether.co.kr/images/homfunctor_preservation_example.svg)

I apologize for taking the liberty to rename objects and morphisms as shown in the diagram above for it was easier to work with for me. I have also taken out all diagonal morphisms and identity morphism minus the one shown for simplicity of proving the preservation rules.

*Unit Preservation*:

So we start with identities \\(1_a:a \rightarrow a\\) and \\(1_b:b \rightarrow b\\) and hope that when we take the homfunctor \\(C(1_a , 1_b)\\), it is the identity for objects in homfunctor, \\(1_{C(a,b)}\\). First take the product \\((1_a, 1_b) = (a \rightarrow a, b \rightarrow b)\\) and then by taking the homfunctor, we get the morphism \\(C(1_a , 1_b) : C(a,b) \rightarrow C(a,b)=1_{C(a,b)}\\) which is too trivial to see the details.

*Composition Preservation*:

We need to show \\(C(f,i) \circ C(g,h)= C(g \circ f, i \circ h)\\). The left hand side is the composition shown in the diagram on the right which takes the object \\(C(a,b) \rightarrow C(a',b) \rightarrow C(a'',b'')\\). On the right side, we get \\(C(g \circ f, i \circ h) = C(a'' \rightarrow a' \rightarrow a, b \rightarrow b' \rightarrow b'') = C(a'' \rightarrow a, b \rightarrow b'')\\) which is just the morphism \\(C(a,b) \rightarrow C(a'',b'')\\) as you can see by the composition \\(i \circ h \circ C(a,b) \circ g \circ f:a'' \rightarrow b''\\).

For the newbies like I, while doing this puzzle found this to be helpful when translating from diagrams to equations.

![homfunctor equation](http://aether.co.kr/images/homfunctor_equation.svg)

![homfunctor preservation rules](http://aether.co.kr/images/homfunctor_preservation_example.svg)

I apologize for taking the liberty to rename objects and morphisms as shown in the diagram above for it was easier to work with for me. I have also taken out all diagonal morphisms and identity morphism minus the one shown for simplicity of proving the preservation rules.

*Unit Preservation*:

So we start with identities \\(1_a:a \rightarrow a\\) and \\(1_b:b \rightarrow b\\) and hope that when we take the homfunctor \\(C(1_a , 1_b)\\), it is the identity for objects in homfunctor, \\(1_{C(a,b)}\\). First take the product \\((1_a, 1_b) = (a \rightarrow a, b \rightarrow b)\\) and then by taking the homfunctor, we get the morphism \\(C(1_a , 1_b) : C(a,b) \rightarrow C(a,b)=1_{C(a,b)}\\) which is too trivial to see the details.

*Composition Preservation*:

We need to show \\(C(f,i) \circ C(g,h)= C(g \circ f, i \circ h)\\). The left hand side is the composition shown in the diagram on the right which takes the object \\(C(a,b) \rightarrow C(a',b) \rightarrow C(a'',b'')\\). On the right side, we get \\(C(g \circ f, i \circ h) = C(a'' \rightarrow a' \rightarrow a, b \rightarrow b' \rightarrow b'') = C(a'' \rightarrow a, b \rightarrow b'')\\) which is just the morphism \\(C(a,b) \rightarrow C(a'',b'')\\) as you can see by the composition \\(i \circ h \circ C(a,b) \circ g \circ f:a'' \rightarrow b''\\).

For the newbies like I, while doing this puzzle found this to be helpful when translating from diagrams to equations.

![homfunctor equation](http://aether.co.kr/images/homfunctor_equation.svg)