Definition
Categories and are equivalent if there exist functors
F:\mathcal{C} \longrightarrow \mathcal{D} \quad \text{and} \quad G: \mathcal{D} \longrightarrow \mathcal{C}$$ and [[202409162200|natural isomorphisms]]\eta: \text{id}\mathcal{C} \Longrightarrow G \circ F \quad \text{and} \quad \varepsilon:F \circ G \Longrightarrow \text{id}\mathcal{D}.
In this case, we call $G$ a **quasi-inverse** of $F$. # Conditions for equivalence A functor $F: \mathcal{C} \to \mathcal{D}$ is part of an equivalence of categories if and only if $F$ is [[202409171519|fully]] [[202409171523|faithful]] and [[202409171525|essentially surjective]]. ^3f61ac ## Proof Assume $F$ is fully faithful and essentially surjective. To prove $\mathcal{C}$ and $\mathcal{D}$ are equivalent, we need to define the "inverse" functor along with the natural isomorphisms. For each object $Z \in \mathcal{D}$, choose an object $X_Z \in \mathcal{C}$ and an isomorphismg_Z: F(X_Z) \longrightarrow Z
in $\mathcal{D}$. These are guaranteed to exist by the essentially surjective condition. Define the functor\begin{align*} G: \mathcal{D} &\longrightarrow \mathcal{C}\ Z &\longmapsto X_Z \ Z_1 \xmapsto{f} Z_2 &\longmapsto X_{Z_1} \xmapsto{G(f)} X_{z_2} \end{align*}
where $G$ is characterized by the requirement that the following diagram commutes: ```tikz \usepackage{tikz-cd} \usepackage{amsmath} \tikzcdset{scale cd/.style={every label/.append style={scale=#1}, cells={nodes={scale=#1}}}} \begin{document} \begin{tikzcd}[scale cd=2, sep=huge] F(X_{Z_1}) \arrow[r, "F(G(f))"] \arrow[d, "g_{Z_1}"] & F(X_{Z_2}) \arrow[d, "g_{Z_2}"] \\ Z_1 \arrow[r, "f"] & Z_2 \end{tikzcd} \end{document} ``` Since $F$ is fully faithful, this completely determines $G(f)$ (you are guaranteed to have one by fullness, and it will be the unique one by faithfulness). Now it remains to construct the natural isomorphisms. Denote them as above\eta: \text{id}\mathcal{C} \Longrightarrow G \circ F \quad \text{and} \quad \varepsilon:F \circ G \Longrightarrow \text{id}\mathcal{D}.
Components of $\varepsilon$ should be of the form\begin{align*} \varepsilon_Z:F(G(Z)) &\longrightarrow Z \ F(X_Z) &\longmapsto Z \end{align*}
so we can use the isomorphism from above $\varepsilon_Z = g_Z$. Thus, the diagram above commuting implies that the isomorphism is natural. We can do a similar thing for $\eta$, where the components should be of the form\eta_X:X \longrightarrow G(F(X)) = X_{F(X)}.
We know using the $g_X$ isomorphisms we have a mapg_{F(X)}:F(X_{F(X)}) \longrightarrow F(X).
Thus, since $F$ is fully faithful, we take $\eta_X$ to be the unique map such that forF(\eta_X):F(X) \longrightarrow F(X_{F(X)})
$F(\eta_X) = g^{-1}_{F(X)}$. This is the more common direction to use this theorem, so I won't include the other direction yet. # Examples ## Finite dimensional vector spaces Let $\text{vect}_k$ be the category of finite dimensional vector spaces over a field $k$ and let $\text{vect}_{k,0}$ be the category of whose objects are the vector spaces $k^n$, $n \geq 0$, and whose morphisms are $k$-linear maps. $\text{vect}_k$ and $\text{vect}_{k,0}$ are equivalent. ### Proof First, we define the functors we will be using. We can see that all $\text{obs}(\text{vect}_{k_0}) \subset \text{obs}(\text{vect}_k)$, so the first functor $F$ is the easier one to define.\begin{align*} F: \text{vect}_{k,0} &\longrightarrow \text{vect}_k \ k^n &\longmapsto k^n\ k^m \xrightarrow{f} k^n &\longmapsto k^m \xrightarrow{f} k_n \end{align*}
For the other way, for every $V \in \text{vect}_k$ chose a basis $\{v_i\}$ and let $\varphi_V \in \text{Hom}_{\text{vect}_k}(V, k^{\dim V})$ be the transformation that maps coefficients given the basis $\{v_i\}$ of the standard basis in $k^{\dim V}$. Then, we can define the functor\begin{align*} G: \text{vect}k &\longrightarrow \text{vect}{k,0}\ V &\longmapsto k^{\dim V}\ V \xrightarrow{f} W &\longmapsto k^{\dim V} \xrightarrow{\varphi_W \circ f \circ \varphi_V^{-1}} k^{\dim W} \end{align*}
Next, to prove equivalence we need to give the natural isomorphisms $\eta: \text{id}_{\text{vect}_{k,0}}\Rightarrow G \circ F$ and $\varepsilon:F \circ G \Rightarrow \text{id}_{\text{vect}_k}$. For each $x \in \text{vect}_{k,0}$, let $\eta_X:X \to X$ be the identity map on $X$. This gives us the following diagram: ```tikz \usepackage{tikz-cd} \usepackage{amsmath} \tikzcdset{scale cd/.style={every label/.append style={scale=#1}, cells={nodes={scale=#1}}}} \begin{document} \begin{tikzcd}[scale cd=2, sep=huge] \text{id}_{\text{vect}_{k,0}}(X) \arrow[r, "\text{id}_{\text{vect}_{k,0}}(f)"] &\text{id}_{\text{vect}_{k,0}}(Y) \\ X \arrow[r, "f"]\arrow[u, phantom, sloped, "="]\arrow[d, "\eta_x = \text{id}_X", swap]& Y\arrow[d, "\eta_Y = \text{id}_Y"]\arrow[u, phantom, sloped, "="]\\ X \arrow[r, "f"]& Y\\ G \circ F(X) \arrow[u, phantom, sloped, "="]\arrow[r, "f"]& G \circ F(Y)\arrow[u, phantom, sloped, "="] \end{tikzcd} \end{document} ``` In this case it is trivial that the diagram commutes as $$\text{id}_Y \circ f = f \circ \text{id}_X$$ Now for the harder one! Let $\varepsilon_X = \varphi_X^{-1}$ with $\varphi_X$ defined as above using the chosen basis of $X$. This gives the following diagram for any $f \in \text{Hom}_{\text{vect}_k}(X, Y)$: ```tikz \usepackage{tikz-cd} \usepackage{amsmath} \tikzcdset{scale cd/.style={every label/.append style={scale=#1}, cells={nodes={scale=#1}}}} \begin{document} \begin{tikzcd}[scale cd=2, sep=huge] F \circ G(X) \arrow[r, "F \circ G(f)"] &F \circ G(Y)\\ k^{\dim X} \arrow[r, "\varphi_Y \circ f \circ \varphi_X^{-1}"]\arrow[u, phantom, sloped, "="]\arrow[d, "\varepsilon_X = \varphi_X^{-1}", swap]& k^{\dim Y}\arrow[d, "\varepsilon_Y = \varphi_Y^{-1}"]\arrow[u, phantom, sloped, "="]\\ X \arrow[r, "f"]& Y\\ \text{id}_{\text{vect}_k}(X) \arrow[u, phantom, sloped, "="]\arrow[r, "\text{id}_{\text{vect}_k}(f)"]& \text{id}_{\text{vect}_k}(Y) \arrow[u, phantom, sloped, "="] \end{tikzcd} \end{document} ``` Starting from the upper left hand corner and going right across the top first then down gives $$\varphi_Y^{-1} \circ \varphi_Y \circ f \circ \varphi_X^{-1} = f \circ \varphi_X^{-1}$$ which is the same as going down first then right along the bottom. Therefore, the diagram commutes and $\varepsilon$ is a natural transformation. Importantly, $\eta_X$ is an isomorphism for each $X \in \text{vect}_{k_0}$ trivially, but $\varepsilon_X = \varphi_X^{-1}$ is also an isomorphism in $\text{vect}_k$ (the inverse is the obvious inverse that must be linear). Hence, $\text{vect}_k$ and $\text{vect}_{k,0}$ are equivalent.