Retraction of a Pushout is a Pushout Suppose we have a diagram with the outer square being a Pushout. Each asterisk is a different object, and morphisms $R,I$ are different morphisms that satisfy $RI=id$.I wish to show that the inner square also satisfies the universal property of pushout.

Note that the diagram may not necessarily be commutative. For example, I think that $IiR=j$ doesn't hold in general. Given two morphisms to some object, we can get a unique arrow from the outer right bottom object using the universal property:

We have that $F_1R=\Phi j$, and the same for bottom arrows some of which I left unnamed. I'm tempted to say that $\Phi Ii=F_1$, if if $\Phi I$ was unique, it would prove that the inner square satisfies the universal property. Since I've mentioned that the diagram may not be commutative, I cannot really conclude $\Phi Ii=F_1$. If it was though, I would claim that $j=IiR$ and from $F_1R=\Phi j=\Phi IiR$ I would get that $F_1=\Phi Ii$ by cancelling $R$ on right. However, I don't think I can do it. Even if I managed to show existence, I also have troubles with showing uniqueness. I think I could let there be another morphism $\Phi^{\prime}$ so that $F_1 = \Phi^{\prime} i$ and somehow show that it must be equal to $\Phi I$. We get an arrow $\Phi^{\prime} R$. Then we have $\Phi^{\prime} R I i R = F_1 R$. If we assume the initial diagram commutes, the latter equality simplifies to $\Phi^{\prime} R j = F_1 R$, which further implies $\Phi^{\prime} R = \Phi$  by the universal property of the outer square. Finally, after composing with $I$ on the right, we imply that $\Phi^{\prime} = \Phi I$.
I think I worked through it correctly, but commutativity is what confused me.

What am I missing?
Motivation. I may be missing some details in my question, so I will include the context. This lemma should be used to prove a corollary to Seifert-van Kampen theorem for subgroupoids.

Corollary. Let $X$ be a topological space covered by open sets $U,V$ with $A$ being its subset. Let $\Pi_A(X)$ be a full (fundamental) subgroupoid with objects being elements of $A$. If $A$ intersects every path component of $U,V, U\cap V$, then the diagram below is a pushout.

Note that we get the usual statement of Seifert-van Kampen for fundamental groupoids if we remove the $A$-subscript. $i, j$ above are just inclusions. The proof goes by constructing left-inverse functors $R$ for each of inclusion functor $I: \Pi_A(U) \to \Pi(U)$ and then using the statement from my question, i.e. by using the diagram without the A-subscript as an outer square and the diagram above as the inner square. $R$ are constructed separately for $U \cap V$, $U \setminus V$, and $V \setminus U$ by choosing for each element $x$ of respective set a path to some element $R(x)$ in $A$; object $x$ is mapped to $R(x)$, and path $\gamma: x\to y$ mapped to some path $R(x) \to R(y)$ obtained by composing with chosen paths $x \to R(x)$ and $y \to R(y)$. For elements in $A$ we choose a constant path. This way we have $R$ is a retraction functor and that $R$'s (and I's) define a morphism of squares.
 A: Let $\mathsf{C}$ be a category with colimits of shape $\mathsf{I}$. A cone under a functor $F:\mathsf{I}\to\mathsf{C}$ can be regarded as a functor $\overline{F}:\mathsf{I}^\triangleright\to\mathsf{C}$ extending $F$, where $\mathsf{I}^\triangleright$ is the category obtained from $\mathsf{I}$ by adjoining a terminal object, which we denote by $\infty$. It suffices to show that if $\overline{F},\overline{G}: \mathsf{I}^\triangleright\to\mathsf{C}$ are functors, $\overline{F}$ is a colimit cone, and $\overline{G}$ is a retract of $\overline{F}$ (in the functor category $[\mathsf{I}^\triangleright,\mathsf{C}]$), then $\overline{G}$ is also a colimit cone. Let $\overline{G'}$ be the colimit cone of $\overline{G}\vert_{\mathsf{I}}$. The universal property of colimits gives us a commutative diagram
$$\require{AMScd}
\begin{CD}
\overline{F} @>\operatorname{id}>>\overline{F}\\
@VVV @VVV \\
\overline{G} @>>> \overline{G'}\\
@VVV @VVV \\
\overline{F} @>\operatorname{id}>>\overline{F},
\end{CD}$$
and we wish to show that the middle horizontal arrow is an isomorphism. This follows from the fact that a retract of an isomorphism is again an isomorphism. (You can prove the last claim for $\mathsf{Set}$ and appeal to the  Yoneda lemma.)
