@QiaochuYuan Then the $1$-image is just the codomain monoid. So I was wrong above, it's not that two of the three images of a monoidal functor coalesce, it's that one of them is just equal to the codomain. Either way, only two are non-trivial.

@Turion The description you gave doesn't define a category! If $f:A\rightarrow B$ and $g:C\rightarrow D$, and we happen to have $FB=FC$ then $Ff$ and $Fg$ would be composable in the image, but we can't define $Fg\circ Ff$ because we want it to be $F(f\circ g)$ but $f\circ g$ isn't defined. I found this very unexpected when I learnt it, categories should have sensible images just like groups or topological spaces or anything else! But when you get used to them the $1$- and $2$-images are very friendly.

@Turion As Qiaochu says, there are definitely two notions of image of a functor $F:\mathcal C\rightarrow D$. The "$2$-image" is defined to be the category $\mathcal C$ with two morphisms quotiented together if they have the same image under $F$. The "$1$-image" is defined to be the full subcategory of $\mathcal D$ spanned by the objects of the form $FX$. I gave what I thought was the generalisation of the $1$-image (though Qiaochu disagrees). By "image" do you instead mean the $2$-image or something else?

@QiaochuYuan Surely two of those three images must be the same. After all monoid homomorphisms don't have two notions of image (or do they?).

The monoidal category $\mathrm{1im}\;F$ has the same objects as $\mathcal C$ and its morphisms $A\rightarrow B$ are $\mathcal D(FA,FB)$. Composition is as defined in $\mathcal D$. Define $\otimes$ as in $\mathcal C$ for objects. On morphisms, use the $\otimes$ in $\mathcal D$ to get something in $\mathcal D(FA\otimes FC,FB\otimes FD)$ and then use the coherence maps of $F$ to translate this into $\mathcal D(F(A\otimes C),F(B\otimes D))$. Then the canonical functors $\mathcal C\rightarrow \mathrm{1im}\;F\rightarrow\mathcal D$ are eso and full-and-faithfull respectively, which is what we want.

Reading this and this has me convinced that the right way to think of the image is "having the objects of $\mathcal C$ and morphisms of $\mathcal D$. I think the following definition works:

Maybe this is the same as what Pace Nielsen calls "$\exists$ elimination" in his answer.

Here's a possible approach: I think the ur-example of a problem where you have to make an arbitrary choice is the following: "Let $A$ be non-empty and let $f:A\rightarrow B$. Is $B$ non-empty?". Clearly the only way to proceed is to choose an arbitrary $a\in A$ an consider $f(a)\in B$. Any other problem involving an arbitrary choice reduces to this one. For example we have a function from the set of finite bases of $V$ to the set of inverses to $V\rightarrow V^{**}$, but to show that $V\rightarrow V^{**}$ actually has an inverse we have to actually pick one of these bases.

@DylanWilson I do like showing off my linear algebra skills! But I think the more general point is that if every example we can think of where it looks like you have to make an arbitrary choice turns out to not in fact need an arbitrary choice, then maybe we should begin to suspect that there aren't any situations where you have to make an arbitrary choice.

@მამუკაჯიბლაძე Good point! On the other hand $\mathcal S(V)$ and $\Lambda(V)$ have canonical inclusions into it (when $\mathrm{char} k=0$), but these inclusions aren't algebra homomorphisms.

There's a canonical quotient map $\mathcal T(V)\rightarrow \mathcal{Cl}_q(V)$. Since $\mathcal T(V)$ is an inner product space, I think that $\mathcal{Cl}_q(V)$ has to be isomorphic to the space orthogonal to the kernel of this map.

If $q$ is an inner product, do $\mathcal T(V)$ and $\mathcal{Cl}_q(V)$ get a canonical inner product? If so we could take the adjoint of the quotient map...

@JohannesHahn I think this makes sense. Imagine your tetrahedron sitting on the table. Pick up the top one of the four mini-tetrahedra and put it onto the table so that its base fits into the triangular hole between the other three. Repeat this process recursively and your Sierpinski tetrahedron will be reduced to a triangle.