**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.

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.