I've attended a category theory class and I remember the professor saying that the object $\mathbb{2}$ represents the forgetful functor in the $\mathbf{Poset}$ category. However, I'm not able to prove this proposition—or I might just misremember it.

To recall some of the definitions:

  • The object $\mathbb{2}$ is defined as the poset with two elements $\left\{x_0, x_1\right\}$ and the relation $x_0 \leq x_1$.
  • An object $A \in \mathbf{C}$ is said to represent a functor $F : \mathbf{C} \rightarrow \mathbf{Set}$ if $F X \cong \mathbf{C}(A, X)$, for any $X \in \mathbf{C}$.
  • The forgetful functor $U : \mathbf{Poset} \rightarrow \mathbf{Set}$ drops the ordering structure: $U((X, \leq)) = X.$

Starting from the definitions, it seems that I have to prove that the cardinality of $X$ is equal to the number of monontone functions from $\{x_0, x_1\}$ to $X$:

$$ U((X, \leq)) \cong \mathbf{Poset}(\mathbb{2}, X) \Rightarrow X \cong \left\{f \vert f : \mathbb{2} \rightarrow (X, \leq)\right\}. $$

Does the above equation hold in general? If not, is there a functor which is represented by the object $\mathbb{2}$?

Note: I am aware that the forgetful functor from $\mathbf{Poset}$ to $\mathbf{Set}$ is represented by the singleton object $\mathbf{1}$.

  • 3
    $\begingroup$ you are right $\bf 1$ represents the forgetful functor; $\bf 2$ doesn't. $\endgroup$ Jan 18, 2018 at 9:47
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    $\begingroup$ By Yoneda, an object representing a functor is unique up to isomorphism. $\endgroup$ Jan 18, 2018 at 9:50
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    $\begingroup$ However, $\mathbf{2}$ represents the "less common" forgetful functor, i.e. the functor $(X, \leq) \mapsto \leq$ $\endgroup$ Jan 18, 2018 at 10:01
  • $\begingroup$ @Max So there isn't a single way to define a forgetful functor. Is the "less common" forgetful functor, let's call it $V$, defined as follows: $V((X, \leq)) = \left\{(x, y) \in X^2 \vert \ x \leq y \right\}$ and $V(f)(x, y) = (f(x), f(y))$? (Sorry for my sloppy notation.) $\endgroup$ Jan 18, 2018 at 10:18
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    $\begingroup$ "[I]s there a functor which is represented by the object 2?" Sure, $\mathrm{Hom}(2, -)$! And yes, speaking about the forgetful functor doesn't technically make sense, although it's often obvious what is meant by it (as you can see, this is one of the cases where it isn't necessarily). $\endgroup$
    – user54748
    Jan 18, 2018 at 11:11

1 Answer 1


To sort of sum up the comments, the short answer to the question in the title is "No", assuming we all agree on what "forgetful functor means".

There are plenty of ways to see this; the one I like the best is to say that $U\cong \mathrm{Hom}(\mathbf{1}, -)$ (where $\cong$ denotes natural isomorphism), thus by the Yoneda lemma, if $U$ were also represented by $\mathbf{2}$, we would have $\mathbf{1}\simeq \mathbf{2}$, which is clearly not the case (of course one can also check, say be direct computation, that $U(X)$ and $\mathrm{Hom}(2,X)$ don't, in general, have the same cardinality).

You also ask "does $\mathbf{2}$ represent any functor ?" . As is mentionned in the comments: obviously yes, it represents $\mathrm{Hom}(\mathbf{2}, -)$. Your question was probably "is there any 'natural functor' represented by $\mathbf{2}$ ?".

To that, the answer is less obvious, but nonetheless quite easy to see, and perhaps a nice exercise for you to do : let $V$ be the functor defined by setting $V((X,\leq_X)) := \leq_X$, and $V(f)((x,y)) := (f(x),f(y))$ (why is this well-defined ?). Then $V$ could also be called a "forgetful functor" (essentially because it's faithful); and it is naturally isomorphic to $\mathrm{Hom}(\mathbf{2}, -)$. I'll let you convince yourself of that fact.

Note however, that this functor is "less forgetful" than $U$ because from $V((X,\leq_X))$ you can find $(X,\leq_X)$, whereas you can't do that from $U((X,\leq_X))$: $V$ doesn't really forget anything.

PS: Note that the fact that $\mathbf{1}$ represents $U$ can be seen by some abstract nonsense: indeed $\mathrm{Id}_{\mathbf{Set}} \cong \mathrm{Hom}_{\mathbf{Set}}(\{\star\}, -)$, so $U \cong \mathrm{Hom}_{\mathbf{Set}}(\{\star\}, U(-))$. However, $U$ has a left adjoint which is just the discrete ordering functor, that sends $X$ to $(X, =_{\mid X})$ (one can easily check this). Call this functor $F$. Then, the previous equation yields $U \cong \mathrm{Hom}_{\mathbf{Poset}}(F(\{\star\}), -)$, and $F(\{\star\}) = \mathbf{1}$, which yields the desired natural isomorphism.

  • $\begingroup$ Thanks a lot for summing up and clarifying the discussion—much appreciated! $\endgroup$ Jan 19, 2018 at 7:51

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