How does bash know about its parent's coprocess in this situation, and why does a shebang line change it?

outer.sh:

ls -l /proc/$$/exe
coproc cat
./inner.sh
kill $!

inner.sh:

ls -l /proc/$$/exe
set | grep COPROC || echo No match found
coproc cat
kill $!

When I run ./outer.sh, this gets printed:

lrwxrwxrwx 1 joe joe 0 Jun 16 22:47 /proc/147876/exe -> /bin/bash
lrwxrwxrwx 1 joe joe 0 Jun 16 22:47 /proc/147879/exe -> /bin/bash
No match found
./inner.sh: line 3: warning: execute_coproc: coproc [147878:COPROC] still exists

Since COPROC and COPROC_PID aren't set in the child, how does it know about the one from the parent to be able to give me that warning? Also, I discovered that if I add #!/bin/bash to the top of inner.sh, or if I call bash ./inner.sh instead of just ./inner.sh from outer.sh, then the warning goes away. Why does this change anything, since it's getting ran with a bash subprocess either way?