How do heteromorphisms arising from constant functors work? I've been trying to understand how to properly reason about heteromorphisms, and I've run into a conceptual problem.
In Mac Lane, Bourbaki, and Adjoints: A Heteromorphic Retrospective, it's stated, on page 13 that colimits arise from the universal mapping property;
$\text{Het}(d, c) \simeq \text{Hom}(\text{colim}\ d, c)$, where $c \in C$ and $d \in C^J$, where $J$ is our diagram.
This means that each heteromorphism from $d$ to $c$ should give rise to some morphism out of the colimit.
According to the nlab page on  profunctors, ordinary functors $F : C \to D$ give rise to the profunctor $D(1, F)(d, c) \simeq D(d, F(c))$ going from $c$ in $C$ to $d$ in $D$.
Here's my actual problem. The interval category has a coproduct (which acts just like disjunction in boolean logic). If all of this is true, then I should be able to derive some of its equations from this perspective. In particular, I should be able to derive all its properties from;
$\text{Het}(
\lambda x.\text{if}(x, a, b), c) \simeq \text{Hom}(a + b, c)$
I can derive some from listing out the functors from $I^\mathbb{B}$ to $I$. There are four obvious ones, $p \mapsto p(0_\mathbb{B})$, $p \mapsto p(1_\mathbb{B})$, $p \mapsto 0_I$, $p \mapsto 1_I$. Each gives rise to some profunctor. My issue rests with the constant functors. Consider the profunctor $D(1, p \mapsto 1_I)(0_I, c) \simeq D(0_I, 1_I)$. This tells us that a heteromorphism from $c$ to $0_I$ exists whenever there is a morphism from $0_I$ to $1_I$, but there always exists such a morphism, by the definition of the interval category. This gives us $\text{Het}(\lambda x.\text{if}(x, a, b), 0_I)$. By the universal mapping property, we then conclude a  $\text{Hom}(a + b,0_I)$, so $a + b = 0_I$ for all $a$ and $b$, since $0_I$ is initial, but this is obviously wrong.
I'm clearly missing something about how to properly reason with heteromorphisms. If anyone can tell me what my issue is, I'd greatly appreciate it.
 A: I think you are mixing up a lot of things in a bad way.
It is true that for each functor $F \colon I^{\mathbb B} \to I$ you can get a profunctor $I(F(-),-) \colon {I^\mathbb{B}}^\text{op} \times I \to \mathbf{Set}$ and you can reguard such profunctor as an $\text{Het}_F(-,-) \colon {I^\mathbb{B}}^\text{op} \times I \to \mathbf{Set}$, that for each functor $p \in \colon I^\mathbb{B} \to I$ and each object $c \in I$ gives you a set of heterogeneus morphisms $\text{Het}_F(p,c)$.
What is not true is that for every such $F$ you get a natural isomorphism 
$$\text{Het}_{F}(p,c) \cong I(\text{colim}\ p,c)\ .$$
In particular this does not work in the case at hand where, if I am not mistaken, you are choosing $F$ to be the constant functor sending every diagram in $I^{\mathbb B}$ to the object $0_I$.
In this case you have the natural isomorphism 
$$\text{Het}_F(p,c) \cong I(0_i, c)$$
(the second profunctor is constant in $p$). 
As you noted clearly for $p=\lambda x. if(x,0,1) \in I^\mathbb{B}$ 
you have that $\text{colim}\ p=1_I \ne 0_I$. This proves exactly that not for every $F$ the profunctor $\text{Het}_F$ allows to define/retrive colimits, you have choose whisely the functor.
Hope this helps.
