Precise meaning of "diagram commutes" in a category theory?

Question

It's weird, but none of introductory category theory books I've seen so far provide rigor definition of "diagram commutes" saying, however they do use such a term extensively.

As far as I was able to deduce from a context, one says that diagram commutes iff there always exists exactly one arrow satifsying somewhat composition. Let's take natural transformation as a simple example: $$G(f) \circ \alpha_A = m = \alpha_{A'} \circ F(f)$$ By saying that such a "square commutes" it is meant that there is unique morphism $m$ equal to the specified compositions.

Is it correct?

Answer 1

You're basically right. A diagram commutes iff every composition of arrows from a given object $X$ to a given object $Y$, via any number of intermediate steps, is the same morphism.

Answer 2

The archetypal commutative diagram is the following: Take four objects $X_1, X_2, Y_1, Y_2$ and morphisms $f_i:X_i\to Y_i$ as well as $g_X:X_1\to X_2, g_Y:Y_1\to Y_2$, i.e. a diagram like this: $$ \require{AMScd} \begin{CD} X_1 @>{g_X}>> X_2 \\ @Vf_1VV @Vf_2VV \\ Y_1 @>>g_Y> Y_2 \end{CD} $$ Then this diagram commutes iff $f_2\circ g_X = g_Y\circ f_1$. In other words, starting at $X_1$, "go right" and "go down" commutes.

The general idea is the same: A diagram commutes iff all possible paths (using the arrows explicitly drawn in the diagram) from one object to another object result in the same morphism. There may be caveats, as detailed by @Tobias in a comment to the other answer.