# How to read commutative diagrams

In algebra, we often consider multiple algebraic objects and structure-preserving maps between them at once. After a while, having to write out something like this gets quite cumbersome:

Let $G$ and $H$ be groups and $\varphi:G \to H$ a group homomorphism. Let $p:G \to G / \ker \varphi$ be the canonical quotient homomorphism, so that $p$ maps each $g \in G$ to the coset $g \ker \varphi$. There exists a unique injective group homomorphism $\bar{\varphi}:G/\ker \varphi \to H$ such that $\varphi = \bar{\varphi} \circ p$.

With the language of diagrams, however, we can write the above as follows:

Our goal in this post is to introduce this language, focusing largely on commutative diagrams.

## 1. Arrows

The simplest diagram is of the form

which is just another way of writing $f:A \to B$. In addition, we stipulate that $f$ be structure-preserving. If $A$ and $B$ are groups, then $f$ is assumed to be a group homomorphism. If $A$ and $B$ are vector spaces over the same field $K$, then $f$ is assumed to be a $K$-linear transform. If $A$ and $B$ are metric spaces (or topological spaces), then $f$ is assumed to be a continuous function.

If $A$ and $B$ carry different structures, then we think of $f$ as a function between sets, unless otherwise specified. In other words, we forget the extraneous structures of both $A$ and $B$ and simply consider them as sets, on which “structure-preserving” maps are simply functions. For example, if $A$ is a group and $B$ a vector space, then the above diagram merely implies that $f$ is a function from $A$ to $B$.

Sometimes, $A$ and $B$ might carry common structures. For example, we assume that $A$ is a group and $B$ is a vector space and write

$A \xrightarrow{\mbox{ }\mbox{ }f\mbox{ }\mbox{ }} B$ in the category of groups.

Here we take $f$ to be a function from $A$ to $B$ such that $f(xy) = f(x) + f(y)$ for all $x,y \in A$. This is because the vector space $B$ can also be thought of as an abelian group with respect to the vector addition operation. Once we consider both $A$ and $B$ as groups, then $f$ ought to be a group homomorphism as per the stipulation in the previous paragraph, which corresponds precisely to the assumption we have just made.

The composition of two structure preserving maps $A \xrightarrow{f} B$ and $B \xrightarrow{g} C$ is denoted as follows:

If we want to label the composition $g \circ f:A \to C$ in the diagram as well, then we can draw an extra arrow:

Note the order of writing. To denote $g \circ f$, we must write $f$ in the first arrow, as $g \circ f(x)$ means input** $x$ to $f$, and then input the output to $g$The orientation of the arrows does not matter, so long as the labels are written down in the correct order. For example, the above diagram can also be written as follows:

We stress that we want every arrow to be structure-preserving. Fortunately, most classes of structure-preserving maps that we care about in mathematics behave nicely under function composition. The composition of two group homomorphisms is a group homomorphism; the composition of two linear transformations is a linear transformation; the composition of two continuous functions is a continuous function.

We conclude this section by introducing three particular kind of arrows. The hook-arrow $\hookrightarrow$ indicates that the structure-preserving map in question is injective, the two-headed arrow $\twoheadrightarrow$ indicates that the structure-preserving map in question is surjective, and $\xrightarrow{\sim}$ indicates that the structure-preserving map is an isomorphism. (Formally, $\hookrightarrow$ and $\twoheadrightarrow$ correspond to mono and epi, respectively, but you needn’t worry about these until you take a graduate course in algebra or topology.) of For example, if $A,B,C,D$ are groups, then the diagram

indicates that $f:A \to B$ is an injective group homomorphism, $g:B \to C$ is a surjective group homomorphism, and $h:C \to D$ is a group isomorphism. Using these arrows is not mandatory: we indicate the extra properties of mappings only when the additional information serves a purpose.

## 2. Commutative Diagrams

What if we have multiple arrows on the same algebraic object? As a motivation, we recall the following result from class:

First isomorphism theorem for groups. Let $G$ and $H$ be groups and $\varphi:G \to H$ a group homomorphism. Let $p:G \to G / \ker \varphi$ be the canonical quotient homomorphism, so that $p$ maps each $g \in G$ to the coset $g \ker \varphi$. There exists a unique injective group homomorphism $\bar{\varphi}:G/\ker \varphi \to H$ such that $\varphi = \bar{\varphi} \circ p$.

A diagrammatic representation of the above statement is as follows; note the arrows that indicate the surjectivity of $p$ and the injectivity of $\bar{\varphi}$, respectively:

An important feature of the above diagram is that there are two way to get from $G$ to $H$ by following the arrows: $G \xrightarrow{\varphi} H$ and $G \xrightarrow{p} G/\ker \varphi \xrightarrow{\bar{\varphi}} H$. Since the first isomorphism theorem establishes that $\varphi = \bar{\varphi} \circ p$, both directions yield the same composite map. Indeed, if we pick $x \in G$, then following the first arrow yields $\varphi(x)$, and following the second sequence of arrows yields $(\bar{\varphi} \circ p)(x)$; both are the same element of $H$.

A diagram is said to be commutative if all sequences of arrows from one algebraic object to another algebraic object yield the same composite map.

We note that the above diagram says nothing about the uniqueness of $\bar{\varphi}$. To remedy this, we modify the diagram as follows:

The dashed arrow

indicates that $\bar{\varphi}$ is the unique structure-preserving map that makes the diagram commute, given all the other arrows in the diagram. Indeed, the first isomorphism theorem states that, given two group homomorphisms $\varphi:G \to H$ and $p:G \to G/\ker \varphi$, there exists a unique group homomorphism $\bar{\varphi}:G/\ker \varphi \to H$ such that $\varphi = \bar{\varphi} \circ p$, viz., the diagram commutes.

Important: the arrow

does not mean that $\bar{\varphi}$ is the unique injective structure-preserving map that makes the diagram commute; it indicates that $\bar{\varphi}$ is the unique structure-preserving map that (1) makes the diagram commute and (2) also happens to be injective.

Just to drive home the point, here is an extension of the first isomorphism theorem for groups:

Universal property of quotient maps. Let $G$ and $H$ be groups, $\varphi:G \to H$ a group homomorphism, and $N$ a normal subgroup of $G$ such that $N \subseteq \ker \varphi$. There exists a unique group homomorphism $\bar{\varphi}:G/N \to H$ such that $\varphi = \bar{\varphi} \circ p$. In other words, $\bar{\varphi}$ is the unique structure-preserving map that makes the following diagram commute:

Yet another way of describing the situation is to say that $\varphi$ factors through $p$. We also remark that $\bar{\varphi}$ is injective if and only if $N = \ker \varphi$.

Note that the above diagram is nearly identical to the diagram for the first isomorphism theorem, except that injectivity is now dropped. Indeed, the first isomorphism theorem is a special case of the universal property of quotient maps.

Exercise.The universal property of quotient maps is extremely general. If you are familiar with the theory of quotient vector spaces, then try to formulate a similar universal property for this context. Try also in the context of quotient sets; here you have to translate the kernel condition into something else, as functions between sets do not have kernels. (See my blog post on naïve set theory for the answer.) Once you figure out a formulation for quotient sets, try writing one down for quotient topological spaces.

You might think after reading the above exercise that diagrams carry a deep meaning, but let me assure you that they do not. While it is incredible that many results in different fields of mathematics take the same shape, the magical nature lies in mathematics itself, not the diagrams. The diagrams are just a language, part of what is affectionately known as abstract nonsense.

## 3. Application: Free Vector Spaces

To illustrate the utility of commutative diagrams, we consider a linear-algebraic example. Recall that a basis of a vector space $V$ (over a field $K$) is a collection $\{v_\alpha : \alpha \in I\}$ of vectors such that every vector in $V$ is a finite $K$-linear combination of vectors in $\{v_\alpha\}$. This suggests that we should be able to create a vector space out of any nonempty set $X = \{x_\alpha : \alpha \in I\}$ by defining “finite linear combinations” of the elements of $X$. Taking a finite linear combination of elements of $X$ over a field $K$ is equivalent to assigning scalars from $K$ to finitely many elements of $X$ and declaring all other elements of $X$ to be zero. In other words:

Let $K$ be a field and $X$ a nonempty set. The free $K$-vector space over$X$ is the set of all functions $f:X \to K$ such that the set

is finite. The vector addition and scalar multiplication are defined as follows:

$(f+g)(x) = f(x) + g(x)$ and $(\lambda f)(x) = \lambda f(x)$ for all $x \in X$.

We write $\mathcal{F}_K(X)$ to denote the free $K$-vector space over $X$.

As a simple example, we take the set $X = \{\mbox{x-axis, y-axis}\}$ and consider the free $\mathbb{R}$-vector space $\mathcal{F}_\mathbb{R}(X)$. Then $\mathcal{F}_\mathbb{R}(X)$ is simply the set of functions that assign a real number to “$x$-axis” and another real number to “$y$-axis”. In other words, $\mathcal{F}_\mathbb{R}(X)$ ought to be linear isomorphic to $\mathbb{R}^2$. And this is true. We define two maps $\mathscr{x},\mathscr{y}:X \to \mathbb{R}$ by setting

$\displaystyle \begin{cases} \mathscr{X}(\mbox{x-axis}) = 1; \\ \mathscr{X}(\mbox{y-axis}) = 0; \end{cases}$ and $\displaystyle \begin{cases} \mathscr{Y}(\mbox{x-axis}) = 0; \\ \mathscr{Y}(\mbox{y-axis}) = 1; \end{cases}$

and define a linear transformation $T:\mathbb{R}^2 \to \mathcal{F}_\mathbb{R}(X)$ by setting $T(a,b) = a\mathscr{X} + b\mathscr{Y}$. It is not hard to check that $T$ is a bijective linear transformation, whence $\mathbb{R}^2 \cong \mathcal{F}_\mathbb{R}(X)$.

We now recall from linear algebra that verifying a property of a linear transformation often amounts to checking it on the basis vectors. Indeed, even the construction of a linear transformation can be done this way. Let $V$ and$W$ be vector spaces over a field $K$ and let $\{v_\alpha : \alpha \in I\}$ be a basis of $V$. A function $\tau:\{v_\alpha:\alpha \in I\} \to W$ can be linearly extended to a linear transformation $T:V \to W$ by declaring that

whenever $\sum_{\alpha \in I_0} \lambda_\alpha v_\alpha$ is a finite linear combination of the basis elements $v_\alpha$. We abstract this property as follows:

Universal property of free vector spaces. Let $K$ be a field and $X$ be a nonempty set. Define the canonical injection map $\iota_X:X \to \mathcal{F}_K(X)$ by setting

Given a function $\tau:X \to W$ into a vector space $W$ over $K$, there exists a unique linear transformation $T:\mathcal{F}_K(X) \to W$ such that $\tau = T \circ \iota_X$. In other words, $T$ is the unique linear transformation that makes the following diagram commute:

Yet another way of describing the situation is to say that $\tau$ factors through $\iota_X$.

Proof. By definition, $\iota(X)$ is a basis of $\mathcal{F}_K(X)$. We define a function $\tau':\iota(X) \to W$ by setting $\tau'(\iota(x)) = \tau(x)$. We now take $T$ to be a linear extension of $\tau'$, whence the commutativity of the diagram follows at once. $\square$

Another intuitive fact from linear algebra is that a function that maps basis elements to basis elements can also be linearly extended to a linear transformation. This can be abstracted as follows:

Universal property of free vector spaces II. Let $K$ be a field and $X$ and $Y$ be nonempty sets. Given a function $\tau:X \to Y$, there is a unique linear transformation $T:\mathcal{F}_K(X) \to \mathcal{F}_K(Y)$ that makes the following diagram commute:

In other words, $T$ is the unique linear transformation such that $T \circ \iota_X = \iota_Y \circ \tau$.

The statement $T$ is the unique linear transformation such that $T \circ \iota_X = \iota_Y \circ \tau$ does not make it clear how the above statement is an abstraction of the property we have discussed. The diagram, however, suggests the following: when we think of $X$ and $Y$ as basis elements of $\mathcal{F}_K(X)$ and $\mathcal{F}_K(Y)$, then we can extend $\tau$ to construct $T$.

More importantly, it is not clear how such a statement should be proved at first sight. With the aid of commutative diagrams, however, we can write down a slick proof.

Proof. Consider first the following diagram, without the map $T$ that we wish to construct:

The composite map $\iota_Y \circ \tau$ is a map from the set $X$ into the vector space $\mathcal{F}_K(Y)$. In particular, the following diagram commutes:

Let’s just focus on the top half of the diagram for now:

By the universal property of free vector spaces, there exists a unique linear transformation $T:\mathcal{F}_K(X) \to \mathcal{F}_K(Y)$ such that the following diagram commutes:

Now, let’s fill in the rest of the diagram:

The bottom half commutes, and the top half commutes, and so the whole diagram commutes. (Check this if you’re not sure.) Moreover, the uniqueness of $T$ depends on $\iota_Y \circ \tau$, which, in turn, is determined uniquely by $\iota_Y$ and $\tau$. Therefore, given $\tau$, $\iota_X$, and $\iota_Y$, we have constructed a unique linear transformation $T$ such that the following diagram commutes:

We have obtained the final diagram by simply dropping the auxiliary diagonal arrow. $\square$

Note that the meat of the proof is contained in one diagram:

Even though we have written out the proof in detail, we didn’t have to: the whole proof is contained in this one diagram.

Once you get comfortable with summarizing proofs with diagrams, you will be ready to tackle proofs by diagram chasing, which is a general technique in algebra and topology that renders many lengthy construction arguments completely trivial.

Thanks to higgsss for corrections!

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