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I'm working through a proof that specifying a left adjoint for a functor $G: \mathcal{D} \to \mathcal{C}$ is equivalent to specifying, for each object $A \in Ob( \mathcal{C})$, an initial object of $(A \downarrow G)$. Here $(A \downarrow G)$ represents the category whose objects are pairs $(B,f)$ with $B \in Ob(\mathcal{D})$, $f: A \to GB$, and with morphisms $(B,f) \to (B',f')$ the morphisms $g:B \to B'$ for which the $(A,GB,GB')$ triangle with morphisms $f,\,f',\,Gg$ commutes; essentially morphisms are what you'd expect them to be.

Now one direction of the proof starts as follows: Let $F: \mathcal{D} \to \mathcal{C}$ be a left-adjoint for $G$. We will show $(FA,\eta_A)$ is an initial object of $(A\downarrow G)$ (where as always $\eta_A:A \to FGA$). Given an object of $(A \downarrow G)$, say $(B,f)$, the triangle

\begin{array}{cc} \,\,\,\,\,\,\,A \\ \\ \eta_A \downarrow & \,\,\,\,\,\,\,\searrow \,{f} \\ \\ GFA & \xrightarrow{Gh} & GB \end{array}

commutes iff the triangle

\begin{array}{cc} \,\,\,\,\,\,\,FA \\ \\ 1_{FA} \downarrow & \,\,\,\,\,\,\,\searrow \,{\bar{f}} \\ \\ \,\,\,\,\,\,FA & \xrightarrow{h} & B \end{array}

commutes. (Apologies as always for my awful diagrams, I tried to use this question to figure out how to draw one but couldn't seem to get many of the answers working, if anyone is keen enough to fix them please feel free.)

Here, $\bar{f}$ represents the corresponding map to $f$ under the bijection from the adjunction. However, I don't understand why these two triangles will each commute if the other does: what have we actually done to get from one triangle to another, applied the adjunction? If so, then I don't see why the adjunction should preserve commutativity; it's just a bijection between the maps, and sure it's natural in the objects it maps between but I don't see why that actually implies shared commutative properties between both sides. So, maybe that isn't what we've done to get from one triangle to the other, and there's some other reason why they both share commutativity or non-commutativity, I don't know. I'd be very grateful if someone could enlighten me.

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2 Answers

up vote 3 down vote accepted

As far as I can work out, to get from the first triangle to the second, you apply the functor $F$, which preserves commutativity because it's a functor, and then you use the adjunction to sort out the $FGh\colon FGFA\to FGB$ that appears at the bottom, like this:

$$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{lll} FA&&\\ \da{F_{\eta_A}}&\searrow^{Ff}&\\ FGFA&\ra{FGh}&FGB\\ \da{\varepsilon_{FA}}&&\da{\varepsilon_B}\\ FA&\ra{h}&B \end{array} $$ (Sorry the diagram isn't ideal, but hopefully it's readable). Then this diagram commutes, the triangle at the top because $F$ is a functor, and the square at the bottom because $\varepsilon$ is a natural transformation. Then noting that $1_{FA}=\varepsilon_{FA}\circ F\eta_A$ and $\overline{f}=\varepsilon_B\circ Ff$, we can use commutativity of this diagram to get:

$$h\circ 1_{FA}=h\circ\varepsilon_{FA}\circ F\eta_A=\varepsilon_B\circ FGh\circ F\eta_A=\varepsilon_B\circ Ff=\overline{f}$$

which is the statement that the second triangle commutes.

For the other direction, you apply $G$ to the second triangle to get:

$$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{lll} GFA&&\\ \da{1_{GFA}}&\searrow^{G\overline{f}}&\\ GFA&\ra{Gh}&FGB \end{array} $$

which commutes because $G$ is a functor, so $G\overline{f}=Gh$. Note also that because $\eta$ is a natural transformation, $GFf\circ\eta_A=\eta_{GB}\circ f$. So we have:

$$ GH\circ\eta_A=G\overline{f}\circ\eta_A=G\varepsilon_B\circ GFf\circ\eta_A=G\varepsilon_B\circ\eta_{GB}\circ F=f $$

which is the statment that the first triangle commutes.

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Many thanks for your answer, it was very well explained (despite Math.SE's fiddly commutative diagrams!) –  Spyam May 15 '12 at 12:23
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This is really just naturality. Consider the following diagram:

$$\begin{array}{ccc} D(FA,FA) & \xrightarrow{\varphi_{A,FA}} & C(A,GFA)\\ \downarrow_{D(FA,h)} & & \downarrow_{C(A,Gh)}\\ D(FA,B) & \xrightarrow{\varphi_{A,B}} & D(A,GB)\\ \end{array}$$ where the $\varphi$ arrows are components of the adjunction. Starting at the upper left corner, chase $1_{FA}$ along this diagram. You get that $$\varphi_{A,B}(h)=Gh\circ\eta_A$$ which implies your commutativity relation, since $\bar{f}=\varphi^{-1}_{A,B}(f)$.

This sort of calculation is very common when dealing with adjoints. When I first studied them, I wrote down the four (or two if you're being economical) diagrams expressing the naturality of the adjunction map and translated them into equations. Those turned out to be somewhat more useful for calculations, especially with units and counits.

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Thanks for your response, I chose Matt's because there was slightly more explanation but they were both good. –  Spyam May 15 '12 at 12:23
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