# Is this equality in a double category true?

Caveat: This is a utterly trivial question from a person who always learned to manipulate diagrams in a double category "from the ground"; I'll be glad even if you simply address me to any source which gives precise rules of transformations for such diagrams (and this explains the "reference-request" tag).

My problem is the following: I wonder if the following equality holds

(where $\alpha\colon G\to G'$, $\beta\colon F\to F'$ are natural transformations between functors, but I think you can "interpret" them as 2-cells somewhere else). Can you help me?

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Could you please better specify the term "equality"? Which compositions should be equal? Each of the subdiagrams involving a single natural transformation commute by definition. – Avitus Jun 12 '13 at 15:55
It's true in a 2-category. Do you really need the double category version? – Zhen Lin Jun 12 '13 at 16:02

Let $F,F': \mathcal C\rightarrow \mathcal D$ and $G,G':\mathcal D\rightarrow \mathcal D'$ be functors between the categories $\mathcal C, \dots, \mathcal D'$. With $\alpha: G\rightarrow G'$ and $\beta:F\rightarrow F'$ we denote natural transformations between them. Then, by definition

$\beta_{F(Y)}\circ F(f)=F'(f)\circ \beta_{X}$

and

$\alpha_{G(Z)}\circ G(g)=G'(g)\circ \alpha_{W}$

for any $X,Y\in\mathcal C$, $W,Z\in\mathbb D$ and morphisms $f:X\rightarrow Y$ in $\mathcal C$, $g:Z\rightarrow W$ in $\mathcal D$.

Each subdiagram involving 1 natural transformation on both sides of your identity commutes by definition. You can use this fact to construct the commutativity of all other subdiagrams and both diagrams in your identity representing commutativity as a way to identify "composition along the upper right corner" with "composition along lower left corner".

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This is exactly what I had in mind. Thank you. – Fosco Loregian Jun 12 '13 at 21:50