The definition of Godement product is given in https://ncatlab.org/nlab/show/Godement+product.
It says that given functors \begin{align*} F_1, G_1 : C \rightarrow D \\ F_2, G_2 : D \rightarrow E \end{align*} and natural transformations \begin{align*} \alpha: F_1 \Rightarrow G_1\\ \beta: F_2 \Rightarrow G_2 \end{align*} the components of the Godement product \begin{equation*} \beta \circ \alpha : F_2 \circ F_1 \Rightarrow G_2 \circ G_1 \end{equation*} is given by the equivalent formulas: \begin{align*} (\beta \circ \alpha)_M = \beta_{G_1 (M)} \circ F_2 (\alpha_M) \\ (\beta \circ \alpha)_M = G_2 (\alpha_M) \circ \beta_{F_1 (M)} \end{align*}
I think I know how these definitions came about; \begin{align*} (F_2 \circ F_1)(M) \xrightarrow{F_2(\alpha_M)}(F_2 \circ G_1)(M) \xrightarrow{\beta_{G_1(M)}}(G_2 \circ G_1)(M) \\ (F_2 \circ F_1)(M) \xrightarrow{\beta_{F_1(M)}}(G_2 \circ F_1)(M) \xrightarrow{G_2(\alpha_M)}(G_2 \circ G_1)(M) \end{align*}
However, I can't see how are they equivalent. Can anyone show me? Thank you.
The way I understand it is visually. Start with the commutative square expressing the natural map in D between F1 and G1. Now project this diagram twice into E, first using F2, second time using G2. The two Squares you get are both commutative (property of functors). Imagine them as two opposite faces of a cube diagram in E. The two Squares are connected by the four other squares of the cube. Because of the existence of beta, we can show that these four squares also commute. Now it is possible to define a natural map by going from say the top left of one Square, and travelling to the bottom left corner of the opposite Square. Because everything commutes this is well defined and gives a natural map.