On

> **Puzzle 18.** Does \\( f_{\ast} \\) always have a left adjoint? If so, describe it. If not, give an example where it doesn't, and some conditions under which it *does* have a left adjoin.

> **Puzzle 19.** Does \\(f_{\ast}\\) always have a right adjoint? If so, describe it. If not, give an example where it doesn't, and some conditions under which it *does* have a right adjoin.

Here is my try:

My current guess is, that the answers depend on whether the original function $$ f : X \to Y $$

was injective, surjective or bijective.

Let's consider 3 cases.

(note 1, \\(PX, PY \\) are posets, not pre-orders, therefore, if there is a left or right adjoint for \\(f_{\ast}\\) -- they would be unique )

(note 2, \\(f_{\ast}\\) is monotone )

1) If \\(f \\) was bijective, then \\(f_{\ast}\\) would be a bijective too.

\\(f_{\ast}\\) would be a bijection, in this case, because:

a) *PY* would have the same number of subsets as *PX*.

b) the image of a subset of PX under f, would be a bijection itself. Therefore, there would be an inverse of that image function. Therefore, the will be an inverse of \\(f_{\ast}\\).

Since there is an inverse, left and right adjoints will be just equal to that inverse of \\(f_{\ast}\\)

Example:

Let's say \\(X=\{1,2\} \\) , \\(f(x) = x+a \\) where a is just some constant natural number.

inverse of f \\(f^{-1}(x) = x - a \\).

\\(PX= \\{1\\},\\{2\\},\\{1,2\\}, \emptyset \\)

\\(PY= \\{1+a\\},\\{2+a\\},\\{1+a,2+a\\}, \emptyset \\)

\\(f_{\ast}\\) would be a monotone function that maps \\( \\{1,2\\} \to \\{1+a,2+a\\} \\) and so on.

It's inverse, is a function that maps back from \\( \\{1+a,2+a\\} \to \\{1,2\\} \\) and so on.

Since the inverse is exact, there are no separate left/right adjoints -- they are all equal to this inverse.

---

2) If \\( f : X \to Y \\) was injective [...deleted as I am rethinking my reasoning ..]

---

3) If \\( f : X \to Y \\) was surjective [...deleted as I am rethinking my reasoning ... ]

Let's say \\(X=\\{1,2,3\\} \\) , \\(f(x) = x \mod 2 \\) . X is cardinality 3, Y is cardinality 2, because there are just 2 members: {0,1}

exact inverse of f does not exist.

\\( PX= \\{1\\}, \\{2\\}, \\{3\\}, \\{1,2\\},\\{1,3\\},\\{2,3\\}, \\{1,2,3\\}, \emptyset \\)

\\(PY= \\{0\\},\\{1\\},\\{0,1\\}, \emptyset \\)

\\(f_{\ast}\\) in this case is also surjective (maps elements of PX onto PY).

Exact inverse of \\(f_{\ast}\\) does not exist, either.

> **Puzzle 18.** Does \\( f_{\ast} \\) always have a left adjoint? If so, describe it. If not, give an example where it doesn't, and some conditions under which it *does* have a left adjoin.

> **Puzzle 19.** Does \\(f_{\ast}\\) always have a right adjoint? If so, describe it. If not, give an example where it doesn't, and some conditions under which it *does* have a right adjoin.

Here is my try:

My current guess is, that the answers depend on whether the original function $$ f : X \to Y $$

was injective, surjective or bijective.

Let's consider 3 cases.

(note 1, \\(PX, PY \\) are posets, not pre-orders, therefore, if there is a left or right adjoint for \\(f_{\ast}\\) -- they would be unique )

(note 2, \\(f_{\ast}\\) is monotone )

1) If \\(f \\) was bijective, then \\(f_{\ast}\\) would be a bijective too.

\\(f_{\ast}\\) would be a bijection, in this case, because:

a) *PY* would have the same number of subsets as *PX*.

b) the image of a subset of PX under f, would be a bijection itself. Therefore, there would be an inverse of that image function. Therefore, the will be an inverse of \\(f_{\ast}\\).

Since there is an inverse, left and right adjoints will be just equal to that inverse of \\(f_{\ast}\\)

Example:

Let's say \\(X=\{1,2\} \\) , \\(f(x) = x+a \\) where a is just some constant natural number.

inverse of f \\(f^{-1}(x) = x - a \\).

\\(PX= \\{1\\},\\{2\\},\\{1,2\\}, \emptyset \\)

\\(PY= \\{1+a\\},\\{2+a\\},\\{1+a,2+a\\}, \emptyset \\)

\\(f_{\ast}\\) would be a monotone function that maps \\( \\{1,2\\} \to \\{1+a,2+a\\} \\) and so on.

It's inverse, is a function that maps back from \\( \\{1+a,2+a\\} \to \\{1,2\\} \\) and so on.

Since the inverse is exact, there are no separate left/right adjoints -- they are all equal to this inverse.

---

2) If \\( f : X \to Y \\) was injective [...deleted as I am rethinking my reasoning ..]

---

3) If \\( f : X \to Y \\) was surjective [...deleted as I am rethinking my reasoning ... ]

Let's say \\(X=\\{1,2,3\\} \\) , \\(f(x) = x \mod 2 \\) . X is cardinality 3, Y is cardinality 2, because there are just 2 members: {0,1}

exact inverse of f does not exist.

\\( PX= \\{1\\}, \\{2\\}, \\{3\\}, \\{1,2\\},\\{1,3\\},\\{2,3\\}, \\{1,2,3\\}, \emptyset \\)

\\(PY= \\{0\\},\\{1\\},\\{0,1\\}, \emptyset \\)

\\(f_{\ast}\\) in this case is also surjective (maps elements of PX onto PY).

Exact inverse of \\(f_{\ast}\\) does not exist, either.