Alright, I am ready to reverse my previous position that \\(A^+\\) is the right/left adjoint of \\(A\\).

In fact, I believe I have found a theorem which makes things a little clearer:

> **Theorem.** Let \\(A : V \to W \\) be a linear map.
> 1. If \\(A\\) is [nonsingular](, then $$ A^{-1}_! \dashv A_! \dashv A^{-1}_! $$ on the category of lattices of subspaces
> 2. If \\(A\\) is singular, then it is impossible for the image of a linear map to be either left or right adjoint to \\(A_!\\)

**Proof.** To prove part 1, observe that \\(A\\) being nonsingular implies it is invertible. We saw in [Puzzle 11]( that this means that \\(A^{-1}_!\\) is both a left and right adjoint.

Next, to prove part 2, assume that \\(A\\) is singular. Then the [kernel]( of \\(A\\) is nonempty. We write this as \\(\mathrm{ker}(A) \neq \varnothing\\). Now assume towards a contradiction there is some linear map \\(R\\) that where \\(A_! \dashv R_!\\). By the characterization of Galois connections I mention in [lecture 5](, this implies \\(X \subseteq (R_! \circ A_!) (X) \\) for all subspaces of \\(X \subseteq V\\). This means in particular that \\(V \subseteq (R_! \circ A_!) (V)\\). Since \\(R_! \circ A_! : V \to V\\), this means \\(V = (R_! \circ A_!) (V)\\). However, we know that \\(R_! \circ A_! = (R \circ A)_!\\). It must be that \\(\mathrm{ker}(R \circ A) = \varnothing\\). But we know \\(\mathrm{ker}(A) \subseteq \mathrm{ker}(R \circ A)\\) and \\(\mathrm{ker}(A) \neq \varnothing\\) by hypothesis, so \\(\mathrm{ker}(R \circ A) \neq \varnothing\\). But then \\(\mathrm{ker}(R \circ A) \neq \varnothing\\) and \\(\mathrm{ker}(R \circ A) = \varnothing\\) simultaneously. ⚡

The proof that \\(A\\) cannot have a left adjoint if it is singular is similar. \\(\Box\\)

I am now starting to suspect

$$ A_! \dashv \mathrm{ker}(A) \vee A^+_! $$

But that might be wishful thinking.