isomorphisms are exactly the morphisms that take colimit cocones to colimit cocones

Question

I would like to understand why isomorphisms are exactly the morphisms that take colimit cocones to colimit cocones. I believe that both directions are easy but they are not immediate to me. Which direction is trivial ? I ask this question just to be safe in preservation of colimits by functors.

For a hint what I mean please see the snippet below; $h$ must be iso:

Answer 1

What's happening is that we have :

• a functor $D\colon \mathcal{D}\to \mathbf{Red}_{\Sigma,\Sigma'}T$
• a colimit cocone $(\delta_d\colon D(d)\to B)_{d\in\mathbf{Ob}(\mathcal{D})}$ in $\mathbf{Red}_{\Sigma,\Sigma'}T$
• a colimit cocone $(\overline{\delta_d}\colon i(D(d))\to A)_{d\in\mathbf{Ob}(\mathcal{D})}$ in $\mathbf{Str}\Sigma$

where $i\colon \mathbf{Red}_{\Sigma,\Sigma'}T \to \mathbf{Str}\Sigma$ denotes the inclusion functor.

Then saying that $i$ preserves colimits means that $(i(\delta_d)\colon i(D(d))\to i(B))_{d\in\mathbf{Ob}(\mathcal{D})}$ is a colimit cocone in $\mathbf{Str}\Sigma$. Since it is a cocone, there is an induced morphism $h\colon A\to i(B)$ with the property that $h\overline{\delta_d}=i(\delta_d)$ for all $d$; and then $(i(\delta_d)\colon i(D(d))\to i(B))_{d\in\mathbf{Ob}(\mathcal{D})}$ is a colimit cocone if and only if $h$ is an isomorphism.

Indeed, if $(i(\delta_d))_d$ is also a colimit cocone, then we can simply use the universal property to get a morphism $h'\colon i(B)\to A$ with the property that $h'i(\delta_d)=\overline{\delta_d}$, and check that $h$ and $h'$ are inverse of one another. Conversely, if $h$ is an isomorphism, and we have some cocone $(\lambda_d\colon i(D(d))\to C)_d$, we get a unique induced morphism $g\colon A\to C$, and then a unique induced morphism $gh^{-1}\colon i(B)\to C$.