A question in definition of product of family in category I am studying categories from Algebra by Thomas Hunger Ford and I have a question in definition of categories :


My question is : What does author means by that diagram is commutative.

I have no clue on what it means although the definition $7.2$ given above is absolutely understood  .
Kindly tell what author means by it.
 A: Basically, what they mean is each triangle in the diagram represents a set of morphism compositions and equalities. For instance,

This particular diagram implies that $\pi_1 \circ \varphi = \varphi_1$. Likewise:

This diagram implies $\pi_2 \circ \varphi = \varphi_1$.
Each of these triangles is considered a commutative diagram, and we also say that the diagram made by "smashing" them together (as you were originally shown) is also commutative.
More generally: in a commutative diagram, whatever paths you take from the same start and end points represent an equality of some sort (in category theory, the equalities concern morphism composition). The first triangle takes two paths from $B$ to $A_1$ for instance: one directly there via $\varphi_1$ and the other goes to $P$ via $\varphi$ and then to $A_1$ via $\pi_1$. Thus, we claim $\pi_1 \circ \varphi = \varphi_1$. Similar occurs for the other diagram, and commutative diagrams in general.
It makes for a nice visual intuition for how these things work, and how equalities can be seen, utilized, and manipulated.
You can find more examples, diagrams, and explanation on the Wikipedia article here.
A: A diagram is commutative iff when we look at all the arrows it generates - that is, all the arrows which can be formed by composing arrows in the diagram itself - we only ever see one arrow between two objects.
For example, suppose we're looking at the category Sets. Consider the objects $A=\{1\}, B=\{2\}, C=\{1,2\}$, and the "triangle" diagram consisting of the arrows $$f:A\rightarrow B: 1\mapsto 2,\quad g: B\rightarrow C: 2\mapsto 2,\quad\mbox{and}\quad h:A\rightarrow C: 1\mapsto 1.$$ This diagram is not commutative: besides the explicitly-present arrows $f,g,h$ themselves, we also have the "generated" arrow $g\circ f$. This has the same domain and codomain as $h$, but is different from $h$.
More snappily:

Commutative triangles are exactly instances of arrows composition: given arrows $f,g,h$ where $g\circ f$ is defined and has the same source and target as $h$, the triangle formed by $f,g,h$ is commutative iff $g\circ f=h$.

There are of course more complicated commutative diagrams out there. Commuting squares crop up frequently (see e.g. "pullback squares"): basically, these correspond to situations where we have arrows $f_1,f_2,f_3,f_4$ such that $f_1$ and $f_2$ have the same source, and $f_3$ and $f_4$ have the same target, and the compositions $$f_3\circ f_1\quad\mbox{and}\quad f_4\circ f_2$$ are (defined and) equal.
