Adjoint to forgetful functor from $\textbf{Met}$ to $\textbf{Set}$

Question

As part of an assignment we need to prove that the forgetful functor $U:\textbf{Met}\to\textbf{Set}$ doesn't have a left adjoint (morphisms in $\textbf{Met}$ are contractions).

For this, I have tried to show that arbitrary sets need not have a free object in $\textbf{Met}$. I have a feeling that as soon as you have more than 1 object in a set, you cannot get the free object; ideally I would like to have this bare-minimum counterexample.

So I have tried multiple ways to prove that $X=\{0,1\}$ doesn't have a free object, but I never really get a general enough result, if that makes sense. I.e. I make assumptions along the way or deal with just specific examples.

E.g. I have tried to give $X$ a metric and have it act as the free object, and that failed the universal property with a specifically chosen test object (since the inclusion map wasn't a contraction). But this only proves that the free object can't have $X$ as its underlying set; no reason it couldn't be a completely different metric space.

I have also tried to smartly choose a test for the UP (including $X$ with a variety of metrics, 3-element sets, etc) to no avail, never really being as generic as possible.

Is there some "easy" counterexample I can use to prove not all sets have a free object in $\textbf{Met}$?

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I do know about solution sets, but for now I would rather avoid that topic, as this is an exercise we should be able to do without that or the adjoint functor theorem.

Answer 1

Let $X = \{0,1\}$ and denote by $X_d$ the metric space with underlying set $X$ and $0$, $1$ having distance $d > 0$. Suppose there exists a free object $F(X)$ of $X$ in $\bf{Met}$, then for every $d > 0$ we get $$\text{Hom}_{\bf{Met}}(F(X),X_d) \cong \text{Hom}_{\bf{Set}}(X,X)$$ which consists of four elements. This implies that $F(X)$ decomposes as $F(X) = U_1 \cup U_2$ with $U_1$ and $U_2$ the connected components of $F(X)$. Indeed, $F(X)$ is clearly disconnected and if $F(X)$ were to decompose as disjoint union of open subsets $U_1$, $U_2$ and $U_3$, an appropriate choice of $d$ (smaller than the minimal distances between elements of $U_1$, $U_2$ and $U_3$) would give us more than four maps $F(X) \rightarrow X_d$.

Now, let $x_i \in U_i$ for $i = 1,2$ and denote by $c$ the distance between $x_1$ and $x_2$. For $d > c$ we get the contradiction that there is a contraction $F(X) \rightarrow X_d$ mapping $x_1$ to $0$ and $x_2$ to $1$.

Answer 2

First, observe that it would have a left adjoint (free metric spaces) if the distance was measured in $[0,+\infty]$ instead of $[0,+\infty)$.

Then, consider the bigger category $\bf F$ that disjointly contains ${\bf Set}$ and ${\bf Met}$ and has functions $X\to M$ from a set to (the underlying set of) a metric space. (By the way, this is the 'collage' of the profunctor $U^*:=\langle X,M\rangle\mapsto\hom_{\bf Set}(X,\,U(M))$ defined by the underlying set functor $U:{\bf Met}\to {\bf Set}$.)

Note that each metric space $M$ has a coreflection to $\bf Set$ in $\bf F$, namely its underlying set $U(M)$, and the left adjoint would mean a reflection for each set onto $\bf Met$.

Now take any arrow (function $f$) from $X=\{0,1\}$ to a metric space $M$, we show that it is not a reflection of $X$.
Set $d:=d(f(0),\,f(1))$ in $M$, and consider $X_{d'}$ with the metric $d(0,1)=d'$ on $X$ where $d'>d$ arbitrary.

Then, the identity function $i:X\to X_{d'}$ cannot factor through $f$, so $f$ was not a reflection.