Forcing Injectivity: From Quotients to Equalizers

Functions are one of the most important mathematical objects, and injectivity is maybe the most important property that a function can have.

Many times you have a function that is not injective and you want to make it injective. In linear algebra and group theory, the solution is to go to the quotient with the morphism’s kernel.

1. Quotient vectorial spaces

When you have a vectorial space V and a subspace U, the quotient space $V/U$ is:

$$V/U = \{v \oplus U: v \in V\}$$

2. Quotient groups

When you have a group G and a normal subgroup H, the quotient $G/U$ group is:

$$G/U =\{ g \cdot H : g \in G \}$$

3. Linear Transformation kernel

$$\begin{gathered} \operatorname{Ker}(T: V \to W) := \{v \in V \mid T(v) = W_{0} \} \end{gathered}$$

4. Forcing injectivity

Every linear transformation $T: V \to W$ induces a linear transformation $T': V/\operatorname{Ker}(T) \to W$ in the following way:

$$T': V/\operatorname{Ker}(T) \to W, \quad v + \operatorname{Ker}(T) \mapsto T(v)$$

This linear transformation is always injective.

4.1 Proof

$$T'(v_1 + \operatorname{Ker}(T)) = T'(v_2 + \operatorname{Ker}(T))$$

Then, by definition of $T'$, we have:

$$T(v_1) = T(v_2)$$

Since $T$ is linear, this implies:

$$T(v_1 - v_2) = 0$$

so $v_1 - v_2 \in \operatorname{Ker}(T)$. because $\operatorname{Ker}(T)$ is indeed a vectorial subspace of V, and addition is closed in this vectorial subspace. Therefore:

$$v_1 + \operatorname{Ker}(T) = v_2 + \operatorname{Ker}(T)$$

Hence $T'$ is injective. $\blacksquare$

The construction above shows that every linear map factors through an injective map. This trick is not only for vectorial spaces; we can do something analogous with groups and normal subgroups. This idea of “forcing injectivity” by identifying elements that differ by elements of the kernel is a fundamental principle that will later generalize to the elegant categorical notion of equalizers.

5. What is an equalizer ?

An equalizer in category theory is a limit of a very special functor. Every categorical diagram can be thought of as a category by completing it with the missing arrows. For example:

$$\begin{array}{ccc} & A & \\ & \swarrow{f} \searrow{g} & \\ B & \xrightarrow{h} & C \end{array}$$

This diagram is not a category because identity arrows are missing, but we can complete this diagram to a category like this:

$$\begin{array}{ccccc} & & A & & \\ & \text{id}_A \nearrow & & \searrow{g} & \\ A & & & & C \\ & \searrow{f} & & \nearrow{\text{id}_C} & \\ & & B & & \\ & & \downarrow{\text{id}_B} & & \\ & & B & & \end{array}$$

This is now a category. Now you can think of every diagram as a category, but a diagram has two parts: points and arrows and the names. Thinking like this, you can think of a diagram as a functor $F: J \to C$ where J is the category of the draw and C is the category where the names live. Basically, when you draw a diagram, you assign to every point and every arrow a name, with some conditions; this is basically what functors do, so you can interpret a diagram as a functor.

An equalizer is the limit of a functor indexed by the following diagram:

$$\bullet \rightrightarrows \bullet$$

A limit object of this diagram can be defined as the universal cone over the functor $F: J \to C$:

$$\overset{A}{\bullet} \mathrel{\substack{\xrightarrow{f} \\ \xrightarrow[g]{}}} \overset{B}{\bullet}$$

A cone with nadir N over this functor has the following property:

$$N \xrightarrow{n} A \mathrel{\substack{\xrightarrow{f} \\ \xrightarrow[g]{}}} B \qquad \text{with } f \circ n = g \circ n$$

And a universal cone with nadir E over this functor has the following property:

This is: $\forall Z \in C$ equipped with a morphism $n: N \to A$ such that
$f\circ n = g \circ n$, there is a unique morphism $k: Z \to E$ such that: $k \circ e = z$

6. Examples of equalizers

Let’s look at two concrete examples. Both kernels we studied before are equalizers: the kernel of a linear transformation $T$ is the equalizer of $T$ and the constant morphism $0$, and the kernel of a group homomorphism $\varphi$ is the equalizer of $\varphi$ and the constant homomorphism $\epsilon$.

6.1 Proof: the kernel of a linear transformation is an equalizer

Let’s think about this equalizer:

First, let’s prove that indeed Ker(T) is an equalizer.

The unique function $k$ is $z$ restricted to $\operatorname{Ker}(T)$; this is:

$$k=z': Z \to \operatorname{Ker}(T), \quad x \mapsto z(x)$$

Suppose that there exists

$$\alpha: Z \to \operatorname{Ker}(T) \text{ such that }i \circ \alpha = z$$

$\text{then } i\circ \alpha = i\circ z'$

$i$ is indeed a monomorphism, then it is left-cancellable; therefore $\alpha = z$. $\blacksquare$

We have seen that T induces an injective linear transformation
$T': V/\operatorname{Ker}(T)) \to W$ because in $V/\operatorname{Ker}(T)$ all elements of $\operatorname{Ker}(T)$ are the same then $\operatorname{Ker}(T') = \{0\}$.

6.2 Proof: the kernel of a group homomorphism is an equalizer

Now let us see the same but with groups. The equalizer is:

The proof that $\operatorname{Ker}(\varphi)$ is indeed an equalizer is analogous to the case of vectorial spaces; we need that the quotient $G/\operatorname{Ker}(\varphi)$ be a group. A sufficient condition for this is: $\operatorname{Ker}(\varphi)$ should be a normal subgroup of $G$. Let’s prove this.

It is a basic result of group theory that $\operatorname{Ker}(\varphi)$ is a subgroup.

Now let $g\in G$ and $k\in\ker(\varphi)$. Then

$$\varphi(gkg^{-1}) =\varphi(g)\varphi(k)\varphi(g^{-1}) =\varphi(g)e_H\varphi(g)^{-1} =e_H.$$

Therefore,

$$gkg^{-1}\in\ker(\varphi).$$

Hence,

$$g\ker(\varphi)g^{-1}\subseteq\ker(\varphi).$$

Replacing (g) by (g^{-1}) gives the reverse inclusion, so

$$g\ker(\varphi)g^{-1}=\ker(\varphi).$$

Thus,

$$\ker(\varphi)\trianglelefteq G.$$

$\blacksquare$

The first isomorphism theorem of group theory guarantees that the induced morphism $\varphi': G/\operatorname{Ker}(\varphi) \to H$ by $\varphi$ is indeed a monomorphism (an injective group homomorphism). This result is also because in $G/\operatorname{Ker}(\varphi)$, all elements of $\operatorname{Ker}(\varphi)$ are the same.

7. Conclusion

Let’s think about the pattern: we have two algebraic structures such that the notion of injectivity for structure-preserving morphisms between objects is equivalent to the condition that the only element of the domain object mapping to the identity element of the codomain object is the identity element of the domain—which, in the case of vector spaces, is the additive identity $V_0$, which is the operation used to construct the quotient. So, when we have structures in a category with equalizers, a notion of injectivity of morphisms characterized by a unique preimage of some element of the codomain of the morphism, quotient objects and the equalizer are suitable for quotients. We can always force injectivity by taking the morphism induced by the original morphism from the quotient of the domain and the equalizer of the original morphism and the constant morphism that sends all elements of the domain to the element that characterizes injectivity in the structure to the codomain of the original morphism.

Also, if we restrict the codomain of the obtained morphism to the image of the original morphism, this new morphism will be a bijection, which is many times a highly desirable property that a morphism can have.

We have seen this phenomenon in vector spaces and groups and it seems like there are a lot of conditions involved, but it is much more common than one might think. but it is far more general. The same construction appears in modules, rings, Lie algebras, Boolean algebras, and, more generally, in every variety of universal algebras. In each case, one quotients the domain by the kernel (or, more fundamentally, by the kernel congruence) to obtain the largest quotient through which the original morphism factors as a monomorphism.

8. References

• Emily Riehl. Category Theory in Context. Aurora: Dover Modern Math Originals. Mineola, NY: Dover Publications, 2016.
• Sheldon Axler. Linear Algebra Done Right. 4th ed. Undergraduate Texts in Mathematics. Cham: Springer, 2024.
• Paolo Aluffi. Algebra: Chapter 0. Graduate Studies in Mathematics, vol. 104. Providence, RI: American Mathematical Society, 2009.