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](http://mathworld.wolfram.com/NonsingularMatrix.html), 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](https://forum.azimuthproject.org/discussion/2055/exercises-and-puzzles-5-chapter-1) 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](https://en.wikipedia.org/wiki/Kernel_(linear_algebra)) 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](https://forum.azimuthproject.org/discussion/comment/16506/#Comment_16506), 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.