**Puzzle 85**

Suppose \$$y\in\mathbb{N}(T)\$$, hence \$$y=a_{E}[E]+a_{Y}[Y]+a_{W}[W]\$$ where \$$a_E, a_Y, a_W\in \mathbb{N}\$$. I am not sure if I am correct but my observation is that in \$$\mathbb{N}(T)\$$, even the reflexive property implies that

$$a_{E}[E]\le a_{E}[E],$$

but we do not have

$$a_{E}[E]\le b_{E}[E],$$

even if \$$a_{E}\le b_{E}\$$ in \$$\mathbb{N}\$$ because we do not have the condition that \$$0\le [E]\$$. \$$0\$$ just serves as the identity in \$$\mathbb{N}(T)\$$ but it does not mean it is less than the "non-zero" elements. This impose strong restrictions on the coefficients on any two "comparable" elements in \$$\mathbb{N}(T)\$$.

Now consider all the \$$x\in\mathbb{N}(S)\$$ in the form \$$x=b_{B}[B]+b_{E}[E]+b_{Y}[Y]+b_{W}[W]+b_{S}[S]\$$ such that \$$y\le f(x)\$$, then we have

$$a_{E}[E]+a_{Y}[Y]+a_{W}[W]\le b_{E}[E]+b_{Y}[Y]+b_{W}[W].$$

In the following, I am considering the case when \$$a_{Y}\le a_{W}\$$, the argument for \$$a_{Y} > a_{W}\$$ will be similar. By the restriction mention above and the relation \$$[Y]+[W]\le [E]\$$ from the reation network, we have

- \$$b_{E}-a_{E}=a_{Y}-b_{Y}=a_{W}-b_{W};\$$
- \$$0\le b_{Y}\le a_{Y};\$$
- \$$a_{W}-a_{Y}\le b_{W}\le a_{W}.\$$

So one would hope for \$$g(y)\in \mathbb{N}(S)\$$ to have the form \$$g(y)=b_{B}[B]+a_{E}[E]+a_{Y}[Y]+a_{W}[W]+b_{S}[S]\$$ because it is supposed to be the "least element" of the set \$$\\left\\{x\in \mathbb{N}(S)| y\le f(x)\\right\\}\$$. So the candidates for \$$g(y)\$$ are

- \$$a_{E}[E]+a_{Y}[Y]+a_{W}[W],\$$
- \$$[B]+a_{E}[E]+a_{Y}[Y]+a_{W}[W],\$$
- \$$a_{E}[E]+a_{Y}[Y]+a_{W}[W]+[S],\$$
- \$$[B]+a_{E}[E]+a_{Y}[Y]+a_{W}[W]+[S],\$$
- etc

However, there is not a common element in \$$\mathbb{N}(S)\$$ that are simultaneously smaller than all the elements above, so I propose that \$$f\$$ does not have a left adjoint.