At first I did not agree with the title "ChatGPT strikes MathOverflow", because we need to verify (somehow) that the answers were indeed created by ChatGPT. I have tested ChatGPT with similar questions like the one about monoidal abelian categories, and although the results were certainly impressive (not correct, but looking good on a superficial level), they were not even close to what the user canvas123 has posted. The answers didn't contain formulas, let alone commutative diagrams.
But then I realized that you can actually add requirements in your query. I asked ChatGPT to give a "detailed proof", which produced formulas as well. Then I asked to "include commutative diagrams as well", which indeed happened (see below).
Interestingly, the line breaks in the commutative diagrams were done with a single \, where a double \\ is required. And the same was the case in the answer by canvas123. Just so that you know, I have edited his diagrams accordingly to make them more readable, because I was curious if the claim that they are nonsensical is correct. After the edit, they were a bit less nonsensical (still wrong).
Given this, it is very likely that canvas123 has indeed used ChatGPT.
It is well-known that ChatGPT often produces wrong answers for complicated, technical questions. Personally, I am still amazed by this tool since sometimes they turn out to be correct, and probably they will even get much better in the next months and years. At some point, it will become very hard to distinguish them from answers written by human beings.
Here is ChatGPT's (wrong) answer to "Write down a detailed proof that every monoidal abelian category admit an exact, lax monoidal functor to abelian groups. Include commutative diagrams as well." (since the modal is stochastic, you will probably get a different answer though):
First, let us construct an abelian category $\mathcal{C}'$ as follows: the objects of $\mathcal{C}'$ are the same as those of $\mathcal{C}$, and the morphisms of $\mathcal{C}'$ are the same as those of $\mathcal{C}$ except that we require them to be strictly additive, i.e. $f + g = h$ implies $f = h-g$ and $g = h-f$. Composition in $\mathcal{C}'$ is the same as in $\mathcal{C}$, and the identity morphism on an object $c$ is the same as the identity morphism in $\mathcal{C}$.
It is straightforward to verify that $\mathcal{C}'$ is indeed an abelian category. To see that it is monoidal, we define the monoidal product $\otimes'$ on $\mathcal{C}'$ as follows: given two objects $c_1, c_2$ in $\mathcal{C}'$, we define $c_1 \otimes' c_2 = c_1 \otimes c_2$ in $\mathcal{C}$. Given two morphisms $f,g$ in $\mathcal{C}'$, we define $f \otimes' g = f \otimes g$ in $\mathcal{C}$. It is straightforward to check that $\otimes'$ is associative and that $I' = I$ is the identity object.
Since $\mathcal{C}'$ is an abelian category, it admits an exact, faithful functor $F: \mathcal{C}' \to \mathbf{Ab}$ to the category of abelian groups. Since $\mathcal{C}'$ is monoidal and $F$ is a functor, it is also lax monoidal, meaning that it preserves the monoidal structure up to natural isomorphism. In other words, there exist natural isomorphisms $$F(c_1 \otimes' c_2) \cong F(c_1) \otimes F(c_2)$$ for all objects $c_1, c_2$ in $\mathcal{C}'$, and $$F(f \otimes' g) \cong F(f) \otimes F(g)$$ for all morphisms $f,g$ in $\mathcal{C}'$. These natural isomorphisms can be represented by the following commutative diagrams: $$\begin{array}{ccc} F(c_1 \otimes' c_2) & \stackrel{\cong}{\longrightarrow} & F(c_1) \otimes F(c_2) \ \downarrow & & \downarrow \ F(c_1) \otimes F(c_2) & \stackrel{\cong}{\longrightarrow} & F(c_1) \otimes F(c_2) \end{array}$$
When you replace \ by \\ in the latter diagram, it looks like this:
$$\begin{array}{ccc} F(c_1 \otimes' c_2) & \stackrel{\cong}{\longrightarrow} & F(c_1) \otimes F(c_2) \\ \downarrow & & \downarrow \\ F(c_1) \otimes F(c_2) & \stackrel{\cong}{\longrightarrow} & F(c_1) \otimes F(c_2) \end{array}$$