Definition (Subcategory):A category $\mathbf{A}$ is a subcategory of a category $\mathbf{B}$ if

Adámek et al,

- $\mathrm{Ob}(\mathbf{A}) \subseteq \mathrm{Ob}(\mathbf{B})$
- For each $A, A' \in \mathrm{Ob}(\mathbf{A})$, $\mathrm{hom}_{\mathbf{A}}(A,A') \subseteq \mathrm{hom}_{\mathbf{B}}(A,A')$.
- For each $A \in \mathrm{Ob}(\mathbf{A})$, the $\mathbf{B}-$identity on $A$ is the same as the $\mathbf{A}-$identity on $A$.

(4) composition in $\mathbf{A}$ is the restriction of composition in $\mathbf{B}$ to the morphisms of $\mathbf{A}$Abstract and Concrete Categories(1990)

Point by point, we'll pick the definition apart. The first part is pretty clear. The collection of objects in a subcategory is contained in the collection of objects in its "parent". The second criterion says that the set of morphisms from one object $A$ to another object $A'$ inside the littler category $\mathbf{A}$ should be a subset of all the morphisms from the same $A$ to the same $A'$, but inside $\mathbf{B}$. That is, there are morphisms from $A \to A'$ in $\mathbf{B}$ that won't live in the subcategory $\mathbf{A}$.

The third criterion just states that the identity morphisms on objects should match in both categories. The final criterion tells us that composition inside the subcategory $\mathbf{A}$ only "works" on the morphisms inside $\mathbf{A}$, but is otherwise the same composition as in $\mathbf{B}$. We just only perform compositions on morphisms in $\mathbf{A}$ when we're in $\mathbf{A}$.

We now define a reflection.

Definition ($A-$ reflection)Let $\mathbf{A}$ be a subcategory of $\mathbf{B}$, and let $B \in \mathrm{Ob}(\mathbf{B})$.

(1) An$\mathbf{A}-$ reflectionfor $B$ is a morphism $B \xrightarrow{r} A$ from $B$ to $A \in \mathrm{Ob}(\mathbf{A})$ such that for any morphism $B \xrightarrow{f} A'$ from $B$ to $A' \in \mathrm{Ob}(\mathbf{A})$ there exists a unique $f': A \to A'$ such that $f = f' \circ r$.

(2) $\mathbf{A}$ is areflective subcategoryfor $\mathbf{B}$ if each $\mathbf{B}-$ object has an $\mathbf{A}-$reflection.

Currently, this definition seems a bit abstract. The following sections will illustrate concrete examples of reflections to understand this definition better.

For this example, we'll be working with the category of groups, $\mathbf{Grp}$. The objects in this category are groups, and the morphisms are group homomorphisms. Some elements of this category are:

- $(\mathbb{R}, +) \xrightarrow{\phi} (\mathbb{R}^{+}, \cdot)$, $\phi(x) = e^{x}$. The real numbers under addition and the positive real numbers under multiplication are both groups, so they're objects in this category. $\phi$ here is a group homomorphism (actually an isomorphism), so it's a morphism in $\mathbf{Grp}$.
- The dihedral group $D_{6}$, and the permutation group $S_{3}$ are also objects in this class. Recall that the dihedral group $D_{6}$ is commonly visualized using the symmetries of an equilateral triangle. (There are 6 symmetry elements. Three reflections, with the reflection lines passing through each of the three vertices, two rotations of angle $2k\pi/3, k=1,2$, and an identity element that does nothing.), More commonly, the dihedral group is given using the following presentation: $$D_{6} = \langle r,s : r^{3} = s^{2} = 1, sr = r^{-1}s\rangle$$ Here, we see that $D_{6}$ is generated by two elements $r$ and $s$ ($r$ is the rotation by $2\pi/3$, and $s$ is one of the reflection lines). $S_{3}$ is the set of permutations on 3 elements-- all permutations of the integers $\{1,2,3\}$. Both of these are groups, and both have 6 elements. If we define $$\phi: \begin{cases} r \to (123) \\ s \to (12)\end{cases}$$ then $\phi$ is a group homomorphism that takes $D_{6}$ to $S_{3}$, and is also thus a morphism in $\mathbf{Grp}$.
- As one last example, let $\mathrm{GL}_{2}(\mathbb{R})$ be the general linear group of degree $2$. This is the group of invertible $2\times 2$ matrices with real entries. Then we can create a group homomorphism $\phi: D_{6} \to \mathrm{GL}_{2}(\mathbb{R})$ by letting $$\phi(r) = \begin{bmatrix}\cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta)\end{bmatrix} \qquad \phi(s) = \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}$$ so this $\phi$ is also a morphism in $\mathbf{Grp}$.

A subcategory of $\mathbf{Grp}$ is $\mathbf{Ab}$, the category of abelian groups. Morphisms are again group homomorphisms, but this time we are only looking at morphisms between abelian groups. Some examples:

- $(\mathbb{R}, +) \xrightarrow{\phi} (\mathbb{R}^{+}, \cdot)$, $\phi(x) = e^{x}$. The real numbers under addition, and the positive real numbers under multiplication are abelian groups, so they're in the subcategory $\mathbf{Ab}$.
- $(\mathbb{Z}_{2} \times \mathbb{Z}_{2}, +)$ and $(\mathbb{Z}_{4}, +)$ are also abelian groups. We can define a group homomorphism $\psi: \mathbb{Z}_{2} \times \mathbb{Z}_{2} \to \mathbb{Z}_{4}$ by $$\psi: \begin{cases} (0,0) \to 0 \\ (1,0) \to 1\\ (0,1) \to 2 \\ (1,1) \to 3\end{cases},$$ so $\psi$ is in the collection of morphisms in $\mathbf{Ab}$. Of course, it's also in the morphisms of $\mathbf{Grp}$ as well.
- $D_{6}$ is not commutative, so it's not an object in $\mathbf{Ab}$.
- $\mathrm{GL}_{2}$ under matrix multiplication is not abelian, because matrix multiplication is not a commutative operation. So neither this group nor the group homomorphism between $D_{6}$ and $\mathrm{GL}_{2}$ are in $\mathbf{Ab}.

We now discuss the commutator element. The** commutator** of two elements $g,h$ in a group, denoted $[g,h]$ is given by $[g,h] = g^{-1}h^{-1}gh$.

*Remark: The order matters here.*

Let's work with a nice easy group of matrices. Let $$G =\left\{I_{2},\begin{bmatrix}0 &1\\1 &0\end{bmatrix},\begin{bmatrix}0 &1\\-1 &-1\end{bmatrix},\begin{bmatrix}-1 &-1\\0 &1\end{bmatrix},\begin{bmatrix}-1 &-1\\1 &0\end{bmatrix},\begin{bmatrix}1 & 0\\-1 &-1\end{bmatrix}\right\}$$ under matrix multiplication. We'll name these matrices $\{I,A,B,C,D,K\}$ respectively. This group is not commutative, as shown in the group table below: \[\begin{array}{c|cccccc} \cdot & I & A & B & C & D & K \\ \hline I & I & A & B & C & D & K \\ A & A & I & C & B & K & D \\ B & B & K & D & A & I & C \\ C & C & D & K & I & A & B \\ D & D & C & I & K & B & A \\ K & K & B & A & D & C & I \\ \end{array} \]

Next, we're going to form the **commutator subgroup**, which is the subgroup of $G$ consisting of all commutators of $G$. Let's call this subgroup $\tilde{G}$. The quotient $G/\tilde{G}$ is always an abelian group, so "quotienting out by" $\tilde{G}$ and working with the cosets gives us an "abelian version" of our previously non-abelian group.

We'll calculate a few commutator elements to demonstrate how, but will not run through every combination. $$\begin{aligned}[I,X] &=[X,I]=I\\ [A,B] &=A^{-1}B^{-1}AB=ADAB=D\\ [B,A] &=B^{-1}A^{-1}BA=DABA=B\\ [C,D] &=C^{-1}D^{-1}CD=CBCD=B\\ \vdots\end{aligned}$$

Continuing with all combinations, we find that there are only three commutator elements: $\{I, B, D\}$. Thus $\tilde{G} = \{I,B,D\}$. Now, $G/\tilde{G}$ gives us the left cosets of $\tilde{G}$: $$\begin{aligned}A\tilde{G} &= C\tilde{G}=K\tilde{G}=\{A,C,K\}\\ B\tilde{G} &= D\tilde{G}=I\tilde{G}=\{I,B,D\}\end{aligned}$$ Thus, the commutator subgroup is $G/\tilde{G} = \{A\tilde{G}, \tilde{G}\}$. This two-element group is a bit dull, admittedly, but it certainly is abelian, with the identity element as $\tilde{G}$.

Something else we can do with this little two-element group is map it to $(\mathbb{Z}_{2}, +)$ via the homomorphism $\phi: G/\tilde{G} \to \mathbb{Z}_{2}$, where $\phi(\tilde{G}) = 0$, and $\phi(A\tilde{G}) = 1$.

What does any of this have to do with reflections? Recall that $G \in \mathrm{Ob}(\mathbf{Grp})$, but not in $\mathrm{Ob}(\mathbf{Ab})$. $G/\tilde{G}$ is in $\mathrm{Ob}(\mathbf{Ab})$, and so is $\mathbb{Z}_{2}$.

A reflection for $G$ into $\mathbf{Ab}$ is a morphism $r$ such that for any morphism $\psi:G \to A'$, $A' \in \mathrm{Ob}(\mathbf{Ab})$, we can find a morphism $\phi$ completely contained in $\mathbf{Ab}$ such that $\psi = \phi \circ r$.

Let $A'= \mathbb{Z}_{2}$, and define $\psi: G \to \mathbb{Z}_{2}$ by $$\psi: \begin{cases}A \to 1 \\ C \to 1 \\ K \to 1 \\ B \to 0\\D \to 0 \\I \to 0\end{cases}$$ This is certainly a homomorphism ($\psi(XY) = \psi(X)\psi(Y)$).

What if we could "bounce" this group $G$ off another group $H$ in $\mathbf{Ab}$, then move from $H$ to $\mathbb{Z}_{2}$? That might reveal a little more ../about $\psi$ than just the seemingly contrived definition we put forth.

Let $r: G \to G/\tilde{G}$ be the canonical map sending each element of $G$ to the appropriate coset. That is $r(A) = A\tilde{G}$, $r(B) = \tilde{G}$, etc. Then the image of $r$ is $G/\tilde{G}$, and $\phi$ as defined above is the unique morphism that will take $G/\tilde{G}$ to $\mathbb{Z}_{2}$ such that $\psi = \phi \circ r$.

One reflection, $r$, taking $G$ to some object $A$ in $\mathbf{Ab}$. Grab a morphism $\psi$ from $G$ to somewhere else in $\mathbf{Ab}$ (we picked $\mathbb{Z_{2}}$). Then we have to be able to find a unique $\phi$ such that $\psi$ decomposes into the composition of $r$ with that $\phi$. This same $r$ (and thus the same $A$) should be able to perform this action for any $A'$, any $\psi$. In our case, the reflection is the canonical map from $G$ to its commutator subgroup.

Reflections can perform many different actions to take objects in one category to objects in a subcategory. We focused on reflections making things "abelian" in a nice way, which helped reveal some structures that would have otherwise been hidden.