Is this correct : whether or not a type 3 grammar generates $\Sigma^*$ is not c.e

Question

An example from Sipser's book, Introduction to the Theory of Computation, shows that it is not decidable for a $TM$ to recognize whether a $CFG$ (or a type 2 grammar) generates $\Sigma^*$, where $\Sigma$ is the alphabet. Call this language $CFG_{all}$

But the above language is also not computably enumerable. There can be a reduction from $CFG_{all}$ to $\bar{A_{TM}}$, where $\bar{A_{TM}}$ is the language s.t. the input $TM$ does not accept any input. $\bar{A_{TM}}$ is not computably enumerable.

But could we say that whether or not a type 3 grammar generates $\Sigma^*$ is also not c.e. ? (since type 3 grammars are a subset of context-free grammars). While it is true that a finite automaton can recognize $\Sigma^*$, this language is different right from whether a type 3 grammar generates $\Sigma^*$?

Just to clarify the source of my confusion, in summary, it is decidable for a $TM$ to decide whether a pushdown automaton recognizes $\Sigma^*$ or accepts any input, but it is not decidable or even computably enumerable for a $TM$ to recognize that a CFG generates $\Sigma^*$. Similarly, it is decidable for a $TM$ to check if a finite automaton accepts $\Sigma^*$, but it may not be decidable for a $TM$ to check if a type 3 grammar generates $\Sigma^*$. It's somehow the difference between recognizing and generating.

EDIT: apparently for a pushdown automaton to recognize $\Sigma^*$ is not decidable

Answer 1

To see if a type 3 grammar (regular) generates $\Sigma^*$ just build the minimal DFA accepting the language, and check if it is just initial = final state, with loop on all symbols. All constructions involved are effective.