[proofplan] We use the universal properties of the two colimit cocones in both directions. The colimit property of $(L,(\lambda_j))$ applied to the cocone $(L',(\lambda'_j))$ gives a unique comparison morphism $\theta: L \to L'$, and the colimit property of $(L',(\lambda'_j))$ applied to $(L,(\lambda_j))$ gives a unique comparison morphism $\psi: L' \to L$. The two composites $\psi \circ \theta$ and $\theta \circ \psi$ are forced to be identity morphisms by the same uniqueness clauses, so $\theta$ is an isomorphism. Finally, the uniqueness of $\theta$ follows directly from the universal property of the first colimit. [/proofplan] [step:Construct the comparison morphism from $L$ to $L'$] Since $(L',(\lambda'_j)_{j \in \operatorname{Ob}(J)})$ is a cocone on $F$ with vertex $L'$, the universal property of the colimit $(L,(\lambda_j)_{j \in \operatorname{Ob}(J)})$ gives a unique morphism \begin{align*} \theta: L \to L' \end{align*} such that \begin{align*} \theta \circ \lambda_j = \lambda'_j \end{align*} for every object $j \in \operatorname{Ob}(J)$. [guided] The first colimit says that every cocone on $F$ receives a unique morphism from $L$ compatible with the cocone maps. We apply this with the target cocone equal to the second colimit cocone \begin{align*} (L',(\lambda'_j)_{j \in \operatorname{Ob}(J)}). \end{align*} Because this is a cocone on $F$, the universal property of $(L,(\lambda_j)_{j \in \operatorname{Ob}(J)})$ produces one and only one morphism \begin{align*} \theta: L \to L' \end{align*} making all structure maps commute: \begin{align*} \theta \circ \lambda_j = \lambda'_j \end{align*} for every object $j \in \operatorname{Ob}(J)$. This is the candidate isomorphism. [/guided] [/step] [step:Construct the comparison morphism from $L'$ to $L$] Since $(L,(\lambda_j)_{j \in \operatorname{Ob}(J)})$ is a cocone on $F$ with vertex $L$, the universal property of the colimit $(L',(\lambda'_j)_{j \in \operatorname{Ob}(J)})$ gives a unique morphism \begin{align*} \psi: L' \to L \end{align*} such that \begin{align*} \psi \circ \lambda'_j = \lambda_j \end{align*} for every object $j \in \operatorname{Ob}(J)$. [guided] Now we use the same idea in the opposite direction. The cocone \begin{align*} (L,(\lambda_j)_{j \in \operatorname{Ob}(J)}) \end{align*} has vertex $L$, and the colimit property of $(L',(\lambda'_j)_{j \in \operatorname{Ob}(J)})$ says that there is a unique morphism from $L'$ to the vertex of any cocone on $F$. Applying that property to this cocone gives a unique morphism \begin{align*} \psi: L' \to L \end{align*} such that \begin{align*} \psi \circ \lambda'_j = \lambda_j \end{align*} for every object $j \in \operatorname{Ob}(J)$. This morphism is the candidate inverse to $\theta$. [/guided] [/step] [step:Show that $\psi \circ \theta$ is the identity on $L$] For each object $j \in \operatorname{Ob}(J)$, associativity of composition in $\mathcal{C}$ gives \begin{align*} (\psi \circ \theta) \circ \lambda_j &= \psi \circ (\theta \circ \lambda_j) \\ &= \psi \circ \lambda'_j \\ &= \lambda_j. \end{align*} Also, \begin{align*} \operatorname{id}_L \circ \lambda_j = \lambda_j \end{align*} for every object $j \in \operatorname{Ob}(J)$. Thus both $\psi \circ \theta: L \to L$ and $\operatorname{id}_L: L \to L$ are mediating morphisms from the colimit cocone $(L,(\lambda_j))$ to the cocone $(L,(\lambda_j))$. By the uniqueness part of the universal property of $(L,(\lambda_j))$, \begin{align*} \psi \circ \theta = \operatorname{id}_L. \end{align*} [guided] To prove that $\psi$ is a left inverse of $\theta$, we compare two morphisms from $L$ to $L$: the composite $\psi \circ \theta$ and the identity morphism $\operatorname{id}_L$. The universal property of the colimit $(L,(\lambda_j))$ will identify them if they induce the same cocone maps. For each object $j \in \operatorname{Ob}(J)$, we compute: \begin{align*} (\psi \circ \theta) \circ \lambda_j &= \psi \circ (\theta \circ \lambda_j) \\ &= \psi \circ \lambda'_j \\ &= \lambda_j. \end{align*} The first equality is associativity of composition in $\mathcal{C}$, the second is the defining property of $\theta$, and the third is the defining property of $\psi$. The identity morphism satisfies \begin{align*} \operatorname{id}_L \circ \lambda_j = \lambda_j \end{align*} for every object $j \in \operatorname{Ob}(J)$. Hence $\psi \circ \theta$ and $\operatorname{id}_L$ are two mediating morphisms from the colimit cocone with vertex $L$ to the same cocone with vertex $L$. The uniqueness clause in the colimit property forces them to be equal: \begin{align*} \psi \circ \theta = \operatorname{id}_L. \end{align*} [/guided] [/step] [step:Show that $\theta \circ \psi$ is the identity on $L'$] For each object $j \in \operatorname{Ob}(J)$, associativity of composition in $\mathcal{C}$ gives \begin{align*} (\theta \circ \psi) \circ \lambda'_j &= \theta \circ (\psi \circ \lambda'_j) \\ &= \theta \circ \lambda_j \\ &= \lambda'_j. \end{align*} Also, \begin{align*} \operatorname{id}_{L'} \circ \lambda'_j = \lambda'_j \end{align*} for every object $j \in \operatorname{Ob}(J)$. Thus both $\theta \circ \psi: L' \to L'$ and $\operatorname{id}_{L'}: L' \to L'$ are mediating morphisms from the colimit cocone $(L',(\lambda'_j))$ to the cocone $(L',(\lambda'_j))$. By the uniqueness part of the universal property of $(L',(\lambda'_j))$, \begin{align*} \theta \circ \psi = \operatorname{id}_{L'}. \end{align*} [/step] [step:Conclude that the comparison morphism is the unique isomorphism] The equalities \begin{align*} \psi \circ \theta = \operatorname{id}_L, \qquad \theta \circ \psi = \operatorname{id}_{L'} \end{align*} show that $\theta$ is an isomorphism with inverse $\psi$. It remains to prove uniqueness among isomorphisms satisfying the stated compatibility. Let $\alpha: L \to L'$ be any morphism such that \begin{align*} \alpha \circ \lambda_j = \lambda'_j \end{align*} for every object $j \in \operatorname{Ob}(J)$. Then $\alpha$ is a mediating morphism from the colimit cocone $(L,(\lambda_j))$ to the cocone $(L',(\lambda'_j))$. The morphism $\theta$ has the same mediating property by construction. By the uniqueness part of the universal property of $(L,(\lambda_j))$, \begin{align*} \alpha = \theta. \end{align*} Therefore $\theta: L \to L'$ is the unique isomorphism satisfying $\theta \circ \lambda_j = \lambda'_j$ for every object $j \in \operatorname{Ob}(J)$. [/step]