Category Theory
Why
- What is category theory? - Mathematics Stack Exchange
- Zermelo–Fraenkel set theory - Wikipedia
- ZFC in nLab
- Class (set theory) - Wikipedia
- Russell’s paradox - Wikipedia
Under ZFC, we cannot construct the special set which contains all sets. We call the set $A$ assuming that such set exists. Let us define a set
\[R := \{ x \in A \mid x \notin x \} .\]This is a valid set by the axiom schema of scpecificaiton. However, since $R \in A$, this implies
\[R \notin R \Leftrightarrow R \in R .\]The existance of such set lead us to Russel’s paradox. The axiom schema of specification has restriciotn that subset itself is not free in $\phi$. But $A$ bypasses this restriction.
To consider such object, and to discuss properties over all sets, category theory has to analyse outsie of ZFC axioms.
Hom sets
- $C, D$,
- category
$C(a, -)$
Covariant hom functor $C(a, -): C \rightarrow \mathrm{Set}$ consists of object funciton
\[\begin{eqnarray} C(a, -): C \rightarrow \mathrm{Set}, \quad C(a, -)(b) & := & C(a, b) \nonumber \end{eqnarray}\]and arrow funciton
\[\begin{eqnarray} f: x \rightarrow y \in C, \quad C(a, -)(f): C(a, x) \rightarrow C(a, y), \quad C(a, -)(f)(g) & := & f \circ g \nonumber . \end{eqnarray}\]$C(-, b)$
Contravariant hom functor $C(-, b): C \rightarrow \mathrm{Set}$ consists of object function
\[\begin{eqnarray} C(-, b): C \rightarrow \mathrm{Set}, \quad C(-, b)(a) & := & C(a, b) \nonumber \end{eqnarray}\]and arrow function
\[\begin{eqnarray} f: y \rightarrow x \in C, \quad C(-, b)(f): C(x, b) \rightarrow C(y, b), \quad C(-, b)(f)(g) & := & g \circ f \nonumber . \end{eqnarray}\]$C^{\mathrm{op}}(a, -)$
Covariant hom functor $C^{\mathrm{op}}(a, -): C^{\mathrm{op}} \rightarrow \mathrm{Set}$ consists of arrow function
\[\begin{eqnarray} C^{\mathrm{op}}(a, -): C \rightarrow \mathrm{Set}, \quad C^{\mathrm{op}}(a, -)(b) & := & C^{\mathrm{op}}(a, b) \nonumber \end{eqnarray}\]and arrow function
\[\begin{eqnarray} f: x \rightarrow y \in C^{\mathrm{op}}, \quad C^{\mathrm{op}}(a, -)(f): C^{\mathrm{op}}(a, x) \rightarrow C^{\mathrm{op}}(a, y), \quad C^{\mathrm{op}}(a, -)(f)(g) & := & f \circ g \nonumber . \end{eqnarray}\]$C(F-, b)$
$F: D \rightarrow C$, Contravariant hom functor $C(F-, b): D \rightarrow \mathrm{Set}$ consists of object function
\[\begin{eqnarray} C(F-, b): D \rightarrow \mathrm{Set}, \quad C(F-, b)(d) & := & C(Fd, b) \nonumber \end{eqnarray}\]and arrow function
\[\begin{eqnarray} f: x \rightarrow y \in D, \quad C(F-, b)(f): C(Fx, b) \rightarrow C(Fy, b), \quad C(F-, b)(f)(g) & := & g \circ Ff \nonumber . \end{eqnarray}\]$C(a, G-)$
$G: D \rightarrow C$, Contravariant hom functor $C(a, G-): D \rightarrow \mathrm{Set}$ consists of object function
\[\begin{eqnarray} C(a, G-): D \rightarrow \mathrm{Set}, \quad C(a, G-)(d) & := & C(a, Gd) \nonumber \end{eqnarray}\]and arrow function
\[\begin{eqnarray} f: x \rightarrow y \in D, \quad C(a, -)(f): C(a, Gx) \rightarrow C(a, Gy), \quad C(a, G-)(f)(g) & := & Gf \circ g \nonumber . \end{eqnarray}\]$C(-, -)$
hom bi-functor $C(-, -): C \times C \rightarrow \mathrm{Set}$ consists of object function
\[\begin{eqnarray} C(-, -): C \times C \rightarrow \mathrm{Set}, \quad C(-, -)(a, b) & := & C(a, b) \nonumber \end{eqnarray}\]and arrow function
\[\begin{eqnarray} f: y \rightarrow x \in C, \quad f^{\prime}: x^{\prime} \rightarrow y^{\prime} \in C, \quad C(-, -)(f, f^{\prime}): C(x, x^{\prime}) \rightarrow C(y, y^{\prime}), \quad g: x \rightarrow x^{\prime}, \quad C(-, -)(f, f^{\prime})(g) & := & f^{\prime} \circ g \circ f \nonumber . \end{eqnarray}\]For all $a \in C$,
\[\begin{eqnarray} f^{\prime}: x^{\prime} \rightarrow y^{\prime} \in C, \quad g: a \rightarrow x^{\prime} \quad C(-, -)(1_{a}, f^{\prime})(g) & = & C(a, -)(f^{\prime})(g) \nonumber \\ f: y \rightarrow x \in C, \quad g: x \rightarrow a \quad C(-, -)(f, 1_{a})(g) & = & C(-, a)(f)(g) \nonumber \end{eqnarray}\]$C(a, -)$ and $C(a^{\prime}, -)$
Let $h: a^{\prime} \rightarrow a$. There is a natural transformation $\tau: C(a, -) \Rightarrow C(a^{\prime}, -)$, $\tau_{x} := C(-, x)(h)$,
\[\begin{CD} x \in C \\ @V{f}VV \\ y \in C \end{CD} \quad \begin{CD} C(a, -)(x) @>{\tau_{x}}>> C(a^{\prime}, -)(x) \\ @V{C(a, -)(f)}VV @V{C(a^{\prime}, -)(f)}VV \\ C(a, -)(y) @>{\tau_{y}}>> C(a^{\prime}, -)(y) \end{CD}\]The equality can be shown by
\[\begin{eqnarray} g \in C(a, x), \quad (C(a^{\prime}, f) \circ \tau_{x})(g) & = & C(a^{\prime}, f)(\tau_{x}(g)) \nonumber \\ & = & f \circ \tau_{x}(g) \nonumber \end{eqnarray}\] \[\begin{eqnarray} g \in C(a, x), \quad (\tau_{y} \circ C(a, f))(g) & = & \tau_{y}(f \circ g) \nonumber \end{eqnarray}\]$C(-, b)$ and $C(-, b^{\prime})$
Let $k: b \rightarrow b^{\prime}$. There is a natural transformation $\tau: C(-, b) \Rightarrow C(-, b^{\prime})$, $\tau_{x} := C(x, -)(k)$,
\[\begin{CD} x \in C \\ @V{f}VV \\ y \in C \end{CD} \quad \begin{CD} C(-, b)(x) @>{\tau_{x}}>> C(-, b^{\prime})(x) \\ @V{C(-, b)(f)}VV @V{C(-, b^{\prime})(f)}VV \\ C(-, b)(y) @>{\tau_{y}}>> C(-, b^{\prime})(y) \end{CD}\]The equality can be shown by
\[\begin{eqnarray} g \in C(x, a), \quad (C(f, b^{\prime}) \circ \tau_{x})(g) & = & C(f, b^{\prime})(\tau_{x}(g)) \nonumber \\ & = & \tau_{x}(g) \circ f \nonumber \end{eqnarray}\] \[\begin{eqnarray} g \in C(x, b), \quad (\tau_{y} \circ C(f, b))(g) & = & \tau_{y}(g \circ f) \nonumber \end{eqnarray}\]$C(a, -)$ and $D(a^{\prime}, G-)$
$G: C \rightarrow D$. Let $\tau: C(a, -) \Rightarrow D(a^{\prime}, G-)$ be a natural transformation.
\[\begin{CD} x \in C \\ @V{f}VV \\ y \in C \end{CD} \quad \begin{CD} C(a, -)(x) @>{\tau_{x}}>> D(a^{\prime}, G-)(x) \\ @V{C(a, -)(f)}VV @V{D(a^{\prime}, -)(f)}VV \\ C(a, -)(y) @>{\tau_{y}}>> D(a^{\prime}, G-)(y) \end{CD}\]The equality can be shown by
\[\begin{eqnarray} g \in C(a, x), \quad (D(a^{\prime}, Gf) \circ \tau_{x})(g) & = & D(a^{\prime}, Gf)(\tau_{x}(g)) \nonumber \\ & = & Gf \circ \tau_{x}(g) \nonumber \end{eqnarray}\] \[\begin{eqnarray} g \in C(a, x), \quad (\tau_{y} \circ C(a, f))(g) & = & \tau_{y}(f \circ g) \nonumber \end{eqnarray}\]$\tau$ is a natural isomorphism if and only if there is a universal arrow $u: a^{\prime} \rightarrow Ga$ from $G$ to $a$. That is, the following are equivalent.
\[\begin{CD} x \in C \\ @V{1_{x}}VV \\ x \in C \end{CD} \quad \begin{CD} C(a, -)(x) @>{\tau_{x}}>> D(a^{\prime}, G-)(x) \end{CD}\]and
\[\begin{CD} a \in C \\ @V{f}VV \\ x \in C \end{CD} \quad \begin{CD} C(a, -)(a) @>{\tau_{a}}>> D(a^{\prime}, G-)(a) \\ @V{C(a, -)(f)}VV @V{D(a^{\prime}, -)(f)}VV \\ C(a, -)(x) @>{\tau_{x}}>> D(a^{\prime}, G-)(x) \end{CD}\]$C(F-, b)$ and $D(-, b^{\prime})$
$F: D \rightarrow C$. Let $\tau: C(F-, b) \Rightarrow D(-, b^{\prime})$ be a natural transformation.
\[\begin{CD} x \in D \\ @V{f}VV \\ y \in D \end{CD} \quad \begin{CD} C(F-, b)(x) @>{\tau_{x}}>> D(-, b^{\prime})(x) \\ @V{C(F-, b)(f)}VV @V{D(-, b^{\prime})(f)}VV \\ C(F-, b)(y) @>{\tau_{y}}>> D(-, b^{\prime})(y) \end{CD}\]The equality can be shown by
\[\begin{eqnarray} g \in C(Fx, a), \quad (D(f, b^{\prime}) \circ \tau_{x})(g) & = & D(f, b^{\prime})(\tau_{x}(g)) \nonumber \\ & = & \tau_{x}(g) \circ f \nonumber \end{eqnarray}\] \[\begin{eqnarray} g \in C(Fx, b), \quad (\tau_{y} \circ C(Ff, b))(g) & = & \tau_{y}(g \circ Ff) \nonumber \end{eqnarray}\]$C(a, -) \circ C(-, b)$
Let $h: x \rightarrow a$, $k: x \rightarrow b$. There is a natural transformation $\tau: C(a, -) \Rightarrow C(-, b)$ defined by $\tau_{x} := C(x, k) \circ C(h, x) = C(h, k)$.
\[\begin{CD} x \in C \\ @V{f}VV \\ y \in C \end{CD} \quad \begin{CD} C(a, -)(x) @>{\tau_{x}}>> C(-, b)(x) \\ @V{C(a, -)(f)}VV @V{C(-, b)(f)}VV \\ C(a, -)(y) @>{\tau_{y}}>> C(-, b)(y) \end{CD}\]Definition
- $C, D$
- category
- $S: D \rightarrow C$,
- functor
- $T: D \rightarrow C$,
- functor
- $\phi: S \Rightarrow T$,
- natural transformation
$\phi$ is said to be natural isomophic if
there is a natural transformation $\phi^{\prime}: T \rightarrow S$ where
\[\phi^{\prime} \circ \phi = 1_{S}, \quad \phi \circ \phi^{\prime} = 1_{T}.\]Here $1_{S}$ is an identity arrow in Functor category.
Reference
- Book: An Introduction to the Language of Category Theory