Duality

If we take a category, we can consider another category where the arrows go in reverse. This is the idea behind the opposite category:

Opposite Category

Let $C$ be a category. The opposite category $C^{\text{op}}$ has

• the same objects as $C$, and
• a morphism $f^{\text{op}}$ for each $f$ in $C$ so that $f^{\text{op}}:X\to Y\in C^{\text{op}}\iff f:Y\to X\in C$

Notice that for $f^{\text{op}}:X\to Y$ and $g^{\text{op}}:Y\to Z$ in $C^{\text{op}}$, we can still compose $g^{\text{op}}{f}^{\text{op}}:X\to Z$ but the composition is in the opposite order in $C$, $fg:Z\to X$.

Any theorem which holds for all categories $C$ also necessarily applies to the opposites of these categories. As a result, any proof in category theory simultaneously proves two theorems: the original, and its dual. For example, consider the following lemma:

Lemma

The following are equivalent

1. $f:x\to y$ is an isomorphism in $C$.
2. For all objects $c\in C$, post-composition with $f$ defines a bijection $f_{\ast }:C(c,x)\to C(c,y)$.
3. For all objects $c\in C$, pre-composition with $f$ defines a bijection $f^{\ast }:C(y,c)\to C(x,c)$.

Proof:

• 1 implies 2: Let $f:x\to y$ have inverse $g:y\to x$. Consider some $h:c\to x$ and $k:c\to y$. Notice that $g_{\ast }{f}_{\ast }(h)=gfh=h$ and $f_{\ast }{g}_{\ast }(k)=fgk=k$. Thus, $f_{\ast }$ is a bijection as $g_{\ast }{f}_{\ast }$ and $f_{\ast }{g}_{\ast }$ are both identities.
• 2 implies 1: Since $f_{\ast }:C(y,x)\to C(y,y)$ is a bijection, there is some $g\in C(y,x)$ such that $f_{\ast }(g)=1_y$. By construction $fg=1_y$. Likewise $f_{\ast }:C(x,x)\to C(x,y)$ being a bijection and $f_{\ast }(gf)=fgf=f$ and $f_{\ast }(1_x)=f$ implies that $gf=1_x$. Thus, $f$ and $g$ are inverses.
• 1 if and only if 3: Notice that we just proved 1 if and only if 2 for all categories. So it follows that $f^{\text{op}}:y\to x$ is an isomorphism if and only if $f_{\ast }^{\text{op}}:C^{\text{op}}(c,y)\to C^{\text{op}}(c,x)$ is a bijection for all $c\in C^{\text{op}}$. This is equivalent to saying that $f^{\ast }:C(y,c)\to C(x,c)$ is a bijection for all $c\in C$ since composition runs backwards.

Mono/epimorphism

A morphism $f:x\to y$ in a category is

1. a monomorphism if for any parallel morphism $h,k:w\to x$, $fh=fk$ implies that $h=k$; or
2. an epimorphism if for any parallel morphism $h,k:y\to z$, $hf=kf$ implies that $h=k$.

In terms of the previous lemma:

1. $f:x\to y$ is a monomorphism in $C$ if and only if for all $c\in C$, $f_{\ast }$ is injective.
2. $f:x\to y$ is an epimorphism in $C$ if and only if for all $c\in C$, $f^{\ast }$ is surjective.

As an example, let $f:X\to Y$ be a monomorphism in the category of sets. Given two maps $x,x^{\prime }:\{c\}\to X$ it follows that $fx=f{x}^{\prime }$ implies that $x=x^{\prime }$. Likewise, if $f:X\to Y$ is an epimorphism, the equation $hf=kf$ implies that the $h=k$ on the image of $f$.

Let $x\stackrel{s}{\to }y\stackrel{r}{\to }x$ be morphisms such that $rs=1_x$. We call $s$ a section or right inverse to $r$ and call $r$ a retraction or left inverse to $s$. In such a case, $s$ is always a monomorphism and $r$ is always an epimorphism. We sometimes call $s$ a split monomorphism and $r$ a split epimorphism.

While isomorphisms are both monic (monomorphism) and epic (epimorphism), there may be morphisms that are both monic and epic but not isomorphisms. For example, the inclusion $f:\mathbb{Z}\hookrightarrow \mathbb{Q}$ is both monic and epic in $\text{Ring}$. Indeed, if $h,k:\mathbb{Q}\to R$ satisfy $hf=kf$ then they are equal on $\mathbb{Z}$. Since they are ring homomorphisms, we get

$$h\left(\frac{x}{y}\right)=h\left(\frac{1}{y}\right)k(x)=f\left(\frac{1}{y}\right)k\left(\frac{xy}{y}\right)=\cdots =k\left(\frac{x}{y}\right).$$

But there is no homomorphisms from $\mathbb{Q}$ to $\mathbb{Z}$ so the morphism is not an isomorphism.

Lemma

1. If $f:x\rightarrowtail y$ and $g:y\rightarrowtail z$ are monomorphisms, then so is $gf:x\rightarrowtail z$.
2. If $f:x\to y$ and $g:y\to z$ are morphisms so that $gf$ is monic; then $f$ is monic.
3. If $f:x\twoheadrightarrow y$ and $g:y\twoheadrightarrow z$ are epimorphisms, then so is $gf:x\twoheadrightarrow z$.
4. If $f:x\to y$ and $g:y\to z$ are morphisms so that $gf$ is epic; then $g$ is epic.

Proof:

1. Let $h,k:w\to x$ such that $gfh=gfk$. Since $g$ is monic, $fh=fk$ and since $f$ is monic, $h=k$. Thus, $gf$ is monic.
2. Let $h,k:w\to x$ such that $fh=fk$. Then clearly $gfh=gfk$, but that implies that $f=k$. Thus, $f$ is monic.
3. Similar to 1.
4. Similar to 2.

Exercises

1. What are the monomorphisms in the category of fields?

Let $f:x\to y$ be a homomorphism. Consider $h,k:w\to x$ such that $fh=fk$. Assume that $h\ne k$. Then there is some $a\in w$ such that $h(a)\ne k(a)$. Notice that

$$0=fh(a)-fk(a)=f(h(a)-k(a))$$

implying that $\text{Ker}f$ is non-trivial. But $\text{Ker}f$ is an ideal, and the only ideals of a field are trivial or the whole field. Thus, $f=0$. Therefore, the morphisms in the category of fields are either trivial or monic.

2. Prove that a morphism that is both a monomorphism and a split epimorphism is necessarily an isomorphism. Argue by duality that a split monomorphism that is an epimorphism is also an isomorphism.

Let $x\stackrel{s}{\to }y\stackrel{r}{\to }x$ such that $rs=1_x$ and $r$ is a monomorphism. Notice that $rsr=1_{x}r=r$. Since $r$ is monic and $rsr=r{1}_y$, $sr=1_y$. Therefore, $r$ is an isomorphism.

Let $x\stackrel{s}{\to }y\stackrel{r}{\to }x$ such that $rs=1_x$ and $s$ is an epimorphism. Notice that by duality $x\stackrel{s^{\text{op}}}{\leftarrow }y\stackrel{r^{\text{op}}}{\leftarrow }x$ satisfies $s^{\text{op}}{r}^{\text{op}}=1_x$ and $s^{\text{op}}$ is a monomorphism. Therefore, $s^{\text{op}}$ is an isomorphism implying that $s$ is an isomorphism.