$\operatorname{Colim}J(k,-)$ is a one point set

Question

I'm working on an excersise from Mac Lane's Categories for the working mathematician, precisely, exercise 4 from page 218 of the second edition which is:

For the covariant Hom-functor $J(k,-):J\to Set$ use the Yoneda lemma to show that $\operatorname{Colim}J(k,-)$ is a one point set.

The chapter being about final functors and $J$ being filtered in the previous exercise, I believe it possible to assume that $J$ is also filtered here. In my attempt of solution I did not assume this.

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Here's what I tried:

The colimit object $C=\operatorname{Colim}J(k,-)$ is nothing other than an universal cone $(J(k,j)\xrightarrow{\lambda_j}C )_{j\in J}$ being exactly a natural transformation $\lambda:J(k,-)\implies C$ where $C$ denotes the constant functor.

The Yoneda bijection states that such natural transformations correspond with elements of the set $C$. I have therefore to show that there is but one such natural transformation, that is, that any other cone $(J(k,j)\xrightarrow{\eta_j}C )_{j\in J}$ is exactly the same one as before. i.e. $\eta_j=\lambda_j$ for all $j\in J$.

Given such a cone, univerality provides us with a unique $\varphi:C\to C$ such that $\varphi \lambda_j =\eta_j$ for all $j\in J$.

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I got stuck here and can't proceed. Is there any reason why $\varphi$ should equal the identity? Is it a property of colimits that all $\lambda_j$ 's are unique? I would appreciate any help.

Answer 1

Let $X$ be a set. Write $\underline{X}$ for the constant functor $\DeclareMathOperator{\Set}{\mathsf{Set}}J\to\Set$ with value $X$. We have the following isomorphisms, natural in $X$: \begin{align*}\DeclareMathOperator{\Hom}{Hom}\DeclareMathOperator{\colim}{colim} \Hom_{\Set}(\varinjlim \Hom_J(k,-),X) &\cong \Hom_{\Set^J}(\Hom_J(k,-),\underline{X})\\ &\cong \underline{X}(k)\\ &\cong X\\ &\cong \Hom_{\Set}(1,X). \end{align*}

The first is the adjunctiom between $\varinjlim$ and $\underline{-}$, the second is Yoneda's Lemma, the third is the definition of the constant functor, and the last is the fact that $1$ represents the identity functor on $\Set$.

By Yoneda again, $\varinjlim \Hom_J(k,-) \cong 1$.

Answer 2

$\newcommand{\J}{\mathcal{J}}$Try to think concretely about what you've been asked. Fix a nonempty locally small category $\J$ and an object $k\in\J$. I can't see a way to hint this without giving the full answer but I'll break this post into multiple spoilers.

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Unless you feel like computing the usually horrendous formula for the colimit, it's rarely useful to appeal to the existence of the colimit set $\varinjlim_\J\J(k,-)$ since you don't (yet) know what that set looks like and can't make the calculations you want to make. Instead, since you've already been told the answer, you should pick a singleton set and check that this is a colimit, that the universal property is really satisfied.

This statement is tantamount to the statement: for any set $X$, every cocone from $\J(k,-)$ to $X$ is constant to an element in $X$ and there is exactly one cocone for every element. Ok, well let's explore why we'd expect cocones to be constant. Firstly, note the second half of this statement follows from the first, since the cocone with constant legs will always exist for a given element of $X$. We just need to show any old cocone will be constant, so fix some cocone $(\phi_j:\J(k,j)\to X)_{j\in\J}$.

Think categorically: the only thing I'm guaranteed to know about is the fact that is $1_k\in\J(k,k)$. It may well be the case no other arrows exist. It's then reasonable to look at the point $p:=\phi_k(1_k)\in X$. I want to show $\phi_j(f)=p$ for any $j$ and any $f:k\to j$. Fix $j$, fix $f$.

How am I going to relate $\phi_j$ to $\phi_k$? Well, the cocone property tells me if I have arrows $k\to j$ or $j\to k$ then composing with $\phi_k,\phi_j$ will give me the same result. More specifically, I know $\phi_j\circ\J(k,g)=\phi_k$ for all arrows $g:k\to j$.

Ok, well it's reasonable to look at $g=f$ then; if I want to relate things to $p=\phi_k(1_k)$, I should probably also evaluate both sides of that equation at $1_k$, right?

So I get: $$(\phi_j\circ\J(k,f))(1_k)=\phi_k(1_k)=p$$But the left hand side is just $\phi_j(f)$, as desired. Done!