Fix a field $k$ and a vector space $W$ over $k$. All other vector spaces will vector spaces over $k$ as well.

Given a linear map $f:V\to V'$, there is a linear map $f^\ast:Hom(V',W)\to Hom(V,W)$ defined by $f^\ast(\alpha)=\alpha\circ f$.

I've been trying to write the details of the proof that $Hom(-,W)$ is a functor $Vect_k^{op}\to Vect$. What confuses me is how to deal with $^{op}$ in proving the functoriality (and even defining $Hom(f,W)$ to begin with).

Let's write $F_W$ for $Hom(-,W)$. It's clear that $F_W$ assigns to each object of $Vect_k^{op}$ an object of $Vect_k$. I need to prove that $F_W$ assigns to each morphism $f: V\to V'$ in $Vect_k^{op}$ a morphism $F_W(f):Hom(V,W)\to Hom(V',W)$ in $Vect_k$ such that $F_W(1_V)=1_{Hom(V,W)}$ and $F_W(g\circ f)=F_W(g)\circ F_W(f)$ whenever $f:V\to V'$ and $g:V'\to V''$ are morphisms in $Vect_k^{op}$.

Let $f: V\to V'$ be a morphism in $Vect_k^{op}$. I know how to assign to it a morphism $Hom(V',W)\to Hom(V,W)$; but I need to assign to it a morphism $F_W(f):Hom(V,W)\to Hom(V',W)$. To this end I need to have a morphism $V'\to V$. But I only have a morphism $f: V\to V'$ in $Vect_k^{op}$. I do understand that it corresponds to a morphism $ V'\to V$ in $Vect_k$, but at this point we are only dealing with $Vect_k^{op}$, not with $Vect_k$. So at least how to define $F_W(f)$ formally, without having a morphism $V'\to V$?


Alright, so one assigns to an arrow $f\in Vect_k^{op}(V,V')=Vect_k(V',V)$ the arrow in $Vect_k(Hom(V,W),Hom(V',W))$ defined by $\alpha\mapsto \alpha\circ f$. But now I still have a problem with functoriality. Suppose $g\in Vect_k^{op}(V',V'')$. The corresponding arrow in $Vect_k(Hom(V',W),Hom(V'',W))$ is defined by $\beta\mapsto \beta\circ g$.

We need to show that $F_W(g\circ f)=F_W(g)\circ F_W(f)$.

Note that $(F_W(g)\circ F_W(f))(\alpha)=F_W(g)(\alpha\circ f)=\alpha\circ f\circ g$.

However, $F_W(g\circ f)(\alpha)=\alpha\circ g\circ f$.

What's wrong?

Update 2:

I think I see what's wrong. It's not true that $F_W(g\circ f)(\alpha)=\alpha\circ g\circ f$. The image of $F_W(g\circ f)$ is the map $\gamma\mapsto \gamma \circ \text{the arrow in the category } Vect_k(V'',V) \text{ corresponding to the arrow } g\circ f \text{ in the category } Vect_k(V,V'')=\gamma\circ f\circ g$

  • $\begingroup$ If $f:V\to V^\prime$ is a morphism in $Vect_k^{op}$ then it is also a morphism $V^\prime\to V$ in $Vect_k$. $\endgroup$ – Theoretical Economist Jun 20 '19 at 1:24
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    $\begingroup$ @user634426 Your second update is correct: $F_W(g \circ_{op} f) = F_W(f\circ g) \equiv [\gamma \mapsto (\gamma \circ f\circ g)] = F_W(g)\circ F_W(f)$ $\endgroup$ – user326210 Jun 20 '19 at 4:32

It can help to distinguish between composition $\circ$ in $\mathrm{Vect}_k$ and composition $\circ_{op}$ in $\mathrm{Vect}_k^{op}$. We can also distinguish by color: $\color{steelblue}{f, g}$ for objects in $\mathrm{Vect}_k$ and $\color{maroon}{f, g}$ for those same objects in $\mathrm{Vect}_k^{op}$.

Then if $f:V\rightarrow V^\prime$ and $g:V^\prime \rightarrow V^{\prime\prime}$, we can interchangeably refer to the composite either as $\color{steelblue}{(g\circ f)}$ in the original category or equivalently as $\color{maroon}{(f\circ_{op} g)}$ in the opposite category.

Functoriality comes from the fact that

$$F_W(\color{maroon}{f})\circ F_W(\color{maroon}{g}) \equiv [\alpha \mapsto (\alpha \circ g)\circ f] = [\alpha \mapsto \alpha \circ \color{steelblue}{(g \circ f)}] = [\alpha \mapsto \alpha \circ \color{maroon}{(f\circ_{op} g)}] = F_W(\color{maroon}{f\circ_{op} g})$$

  • $\begingroup$ But what's wrong with my updated "proof"? I think I started with the opposite category right away. And one of my results ($F_W(g\circ f)(\alpha)=\alpha\circ g\circ f$) is the same as yours (I used $F_W$ instead of your "star notation"). But $F_W(g)\circ F_W(f)(\alpha)$ doesn't match with that result, even though I regarded $f$ and $g$ as arrows in the opposite category. $\endgroup$ – user634426 Jun 20 '19 at 2:39
  • $\begingroup$ The bug is that we have to distinguish two different notions of composition: $F_W(g) \circ F_W(f) = [\alpha \mapsto \alpha \circ f \circ g]$. And $F_W(g\circ_{op} f) \equiv F_W(f\circ g) = [\alpha \mapsto \alpha \circ (f\circ g)] = [\alpha \mapsto \alpha\circ f \circ g]$. See updated answer. $\endgroup$ – user326210 Jun 20 '19 at 3:45

I'm going to repeat some of what you wrote with more words to get myself up to speed.

Let's write $V^{op}$ for the object in ${Vect}_k^{op}$ that corresponds to the vector space $V$ and $f^{op}$ for the arrow in ${Vect}_k^{op}$ corresponding to the linear map $f$.

So, given $f^{op}\colon B^{op} \to A^{op}$, $F_W(f^{op}) = Hom(f,W)$ should be an arrow $Hom(B,W) \to Hom(A,W)$. So how do I take a linear map $\alpha\colon B\to W$ and turn it into one $A \to W$? Well, precompose with $f\colon A \to B$, the linear map that $f^{op}$ corresponds to!

Now how should composition work? The fact that you're working with opposite categories tells me that when we say "functor" we really only ever mean "covariant functor." So it should be true that $F_W(f^{op}g^{op}) = F_W(f^{op})F_W(g^{op}) = F_W([gf]^{op}) = Hom(gf,W) = Hom(g,W)\circ Hom(f,W)$.

And indeed, if given $g \colon B \to C$ and $\alpha\colon C \to W$, I precompose $(\alpha g)\colon B \to W$ with $f\colon A \to B$, I get a morphism $\alpha gf$ in $Hom(A,W)$.

Note that it is emphatically not the case that both compositions make sense. This is why I have insisted on superscripting "op" everywhere just for clarity.

  • $\begingroup$ Passing to the next step, if there are $f\in Vect_k^{op}(V,V')=Vect_k(V',V)$ and $g\in Vect_k^{op}(V',V'')=Vect_k(V'',V')$, how do I know which of the two equalities to prove (to establish functoriality): $F_W(g\circ f)=F_W(g)\circ F_W(f)$ or $F_W(f\circ g)=F_W(f)\circ F_W(g)$? Both $f\circ g$ and $g\circ f$ make sense as compositions. $\endgroup$ – user634426 Jun 20 '19 at 1:38
  • $\begingroup$ I have updated my answer with some thoughts. $\endgroup$ – Rylee Lyman Jun 20 '19 at 1:53
  • $\begingroup$ Do you mean that $g^{op}\in Vect_k^{op}(V',V'')$ and $f^{op}\in Vect_k^{op}(V,V')$? If so, then how does the composition $f^{op}\circ g^{op}$ make sense? I've added some details in the question as well. $\endgroup$ – user634426 Jun 20 '19 at 1:59
  • $\begingroup$ I mean that $f\colon V \to V'$ and $g\colon V' \to V''$, so that $gf \colon V \to V''$. Thus $(gf)^{op} \colon V''^{op} \to V^{op}$. Indeed $g^{op}\colon V''^{op} \to V'{op}$, and $f^{op}\colon V'^{op} \to V^{op}$, so $(gf)^{op} = f^{op}\circ g^{op}$. $\endgroup$ – Rylee Lyman Jun 20 '19 at 2:27
  • $\begingroup$ If $f:V\to V'$ and $g:V'\to V''$, then it makes sense. But your answer says that $f:V'\to V$ and $g: V''\to V'$. $\endgroup$ – user634426 Jun 20 '19 at 2:33

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