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Commutative diagram

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This very fruitful approach is used throughout.

Suppose we are given a function $f \colon X \rightarrow Y$.

We can represent it diagrammatically as: $$ \newcommand{\ra}[1]{\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccccccccccc} input & & {function} & & {output} \\ x & \ra{} & \stackrel{}{\left[...f...\right]} & \ra{} & y \end{array} $$ or $$ \newcommand{\ra}[1]{\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} X & \ra{f} & Y \end{array} $$

The first diagram might come from computer science/programming and the second is the way we do it in algebra. They are very similar...

What if we have another function $g \colon Y \rightarrow Z$, how do we represent their composition?

To compute it, we wire their diagrams together consecutively: $$ \newcommand{\ra}[1]{\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} X & \ra{f} & Y & \ra{g} & Z \\ \end{array} $$

Alternatively, we might emphasize the resulting composition function: $$ \newcommand{\ra}[1]{\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} & X & \ra{f} & Y \\ & _{g \circ f} & \searrow & \da{g} \\ & & & Z \end{array} $$

One can see the composition function that goes diagonally.

The point of the diagram is that you can

  • go right then down, or
  • go diagonally.

Either way, you get the same result: $$g(f(x))=(g \circ f)(x).$$

Specifically, this is how the values of the functions are related: $$ \newcommand{\ra}[1]{\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} & x \in X & \ra{f} & Y \ni f(x) \\ & _{g \circ f} & \searrow & \da{g} \\ & (g \circ f)(x) & \in & Z \ni g(f(x)) \end{array} $$


As an example, we can use this idea to represent the inverse function. We just choose $g=f^{-1}$ above: $$ \newcommand{\ra}[1]{\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} & X & \ra{f} & Y \\ & _{id_X} & \searrow & \da{f^{-1}} \\ & & & X \end{array} $$

More generally, the diagram may be a square, or any other shape: $$ \newcommand{\ra}[1]{\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{cccccccccc} X & \ra{f} & Y \\ \da{g'} & \searrow & \da{g} \\ X' & \ra{f'} & Y' \end{array} $$

Again, go right then down, or go down then right, same result: $$g \circ f = f' \circ g'.$$ Both give you the diagonal function!

This identity is the reason why it makes sense to call it a commutative diagram.

Commutative diagram.png

Also, the above square diagram is commutative in the sense that

vertical then horizontal is same as horizontal then vertical.

The image above explains how the blue and green threads are tied together in the beginning -- as we start with the same $x$ in the left upper corner -- and at the end -- where the output of these compositions in the right bottom corner turns out to be the same. In a sense, the commutativity turns this combination of loose threads into a piece of fabric!

In our study of algebraic topology we encounter large commutative diagrams, such as this: $$ \newcommand{\ra}[1]{\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccccccccc} ...& \to & C_{k+1}(K)& \ra{\partial_{k+1}} & C_{k}(K)& \ra{\partial_{k}} & C_{k-1}(K)& \to&...& \to & 0\\ ...& & \da{f_{k+1}}& & \da{f_{k}}& & \da{f_{k-1}}& &...& & \\ ...& \to & C_{k+1}(L)& \ra{\partial_{k+1}} & C_{k}(L)& \ra{\partial_{k}} & C_{k-1}(L)& \to&...& \to & 0. \end{array} $$ The paths may be long consisting of many arrows. One only needs to verify the commutativity of each square to conclude that any two paths with the same beginning and the same end produce the same result.