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> **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.

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

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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.