Bash trap wrong $BASH_COMMAND

Question

I created two traps

trap function1 DEBUG
trap pwd ERR

function function1 {
echo $BASH_COMMAND
}

If I use some wrong command let's say a, then output is->

a
bash: a: command not found
a
/home/user/dir

So second time DEBUG is called to execute pwd, but why $BASH_COMMAND is still a? Is there any way to know in the function1 that this call to function is made while executing command in trap ERR?

EDIT

Two traps are

trap -- 'f' DEBUG
trap -- 's' ERR

function f 
{ 
    echo "inside function f command is $BASH_COMMAND"
}
function s 
{ 
    echo "inside function s command is $BASH_COMMAND"
}

Output for wrong command a, is

inside function f command is a
bash: a: command not found
inside function f command is a
inside function s command is echo "inside function s command is $BASH_COMMAND"

The second inside function f command is a, must be from ERR trap.

Answer 1

This can be found in the bash documentation (see your man bash), where it writes in several places about various aspects of trap.

BASH_COMMAND The command currently being executed or about to be executed, unless the shell is executing a command as the result of a trap, in which case it is the command executing at the time of the trap.

and also

the DEBUG and RETURN traps [...] are not inherited unless the function has been given the trace attribute [...] or the -o functrace shell option has been enabled with the set builtin (in which case all functions inherit the DEBUG and RETURN traps)

If you modify your trap commands slightly you can see this more clearly

trap 'echo "##DEBUG## $BASH_COMMAND"' DEBUG
trap 'echo "##ERR## $PWD"' ERR

a
##DEBUG## a
-bash: a: command not found
##DEBUG## a
##ERR## /home/roaima

and

f() { printf "%s..." "We are in function f()"; sleep 1; printf " OK\n";  }
f
##DEBUG## f
We are in function f()... OK