What is the "opposite" of a forgetful functor?

Question

Consider a category $C$ and a monoid $M$. Consider a functor $F:C\to M$. It maps the objects of $C$ into the only object of $M$. But I don't want it to map every morphism of $C$ into the identity on $M$. If $f$ is a morphism in $C$, I would like in general $F(f)$ to be non-trivial.

Intuitively, this functor forgets only the underlying graph, and preserves (some of) the operations. In a way, it is the opposite of a forgetful functor. Does a functor with such properties have a name?

Answer 1

There is a construction of a "universal morphism" in Brown's Topology and Groupoids, chapter 8.1. We assume that $G$ is a groupoid, $\sigma:\text{Ob}(G)\to X$ is a set map. Then we can construct a groupoid $U$ whose object set is exactly $X$, and a morphism $\barσ:G\to U$ whose object function is $σ$. The idea is similar to the construction of the free product of groups. We form words of different lengths, where a word of length $n$ from $x$ to $x'$ is a tuple $$a=(a_n,...,a_1)$$ such that each $a_i$ is in $G(x_i,x_i')$ and
(a) $ x_i'\ne x_{i+1}$
(b) $σx_i'=σx_{i+1}$
(c) $σx_i=x,\ σx_n'=x'$
(d) no $a_i$ is the identity

Recall how we form a word in the free product of groups $G_i$. We write $a_1a_2...a_n$ where no adjacent elements can be multiplied (which means they come form different $G_i$), and also no $a_1$ is the identity.

We multiply two words by putting them end to end, composing in $G$ and cancelling identities whenever possible.

The resulting groupoid $U_σ(G)$ has as arrows $x\to x'$ words of arbitrary length, where a word of length $0$ is used as identity. If $x\in X\setminusσ(\text{Ob}(G))$, then the only word starting or ending at $x$ is $\mathbf 1_x$, the identity.

the morphism $\barσ:G\to U_σ(G)$ sends $a\ne \mathbf 1$ to $(a)$ and $a=\mathbf 1_{x_1}$ to $()_{σx_1}$

This groupoid has the following universal property:
If $g:G\to K$ is a morphism whose object function factors as $\text{Ob}(g)=\tauσ$, then there is a unique morphism $g^*:U_σ(G)\to K$ such that $g^*\barσ=g$ and $\text{Ob}(g^*)=\tau$.

Compare this with the definition of the final topology. If $(Y,\mathcal O_Y)$ is a topological space, $X$ is a set, and $s:Y\to X$ is a set map, then there is a topology $\mathcal O_X$ on $X$ making $s$ continuous. For every continuous map $g:Y\to Z$ whose underlying set map factors as $g=\tau s$, there is a unique continuous map $g^*:X\to Z$ such that $g^*s=g$ and $g^*=\tau$ as a set map.

Some interesting cases are when $X=\{*\}$ is a singleton and $σ:\text{Ob}(G)\to X$ the constant function. In that case we obtain a group $UG$ which is universal among all morphisms from $G$ to groups. This is in fact a left adjoint to the inclusion $\mathbf{Grp}\to\mathbf{Grpd}$.

The statements and proofs in that chapter carry over almost verbatim to the case of categories instead of groupoids, and monoids instead of groups.

Coming back to my remark about the free product of groups $(G_i)_{i\in I}$, the universal group-functor $U:\mathbf{Grpd}→\mathbf{Grp}$ is left adjoint to the inclusion, so it preserves coproducts. Now the coproduct of $(G_i)_I$ is $\mathbf{Grpd}$ is their disjoint union $\coprod_I G_i$, while the coproduct in $\mathbf{Grp}$ is the free product $\mbox{*}_IG_i$. We then have $$U\left(\coprod_I G_i\right)=\mbox{*}_I G_i$$ So this gives us the well-known description of the free product of groups. The property (a) then translates to "Adjacent letters of a word must be from distinct groups". Property (b) and (c) are automatically satisfied, as all objects are identified.