Yes the executed process will **not** change the working directory of the parent process.
Example:
* caller.sh
#!/bin/bash
echo -n "Caller 1 " ; pwd
./callee.sh
echo -n "Caller 2 " ; pwd
* callee.sh
#!/bin/bash
echo -n "Callee 1 " ; pwd
cd /tmp
echo -n "Callee 2 " ; pwd
exit
Calling `caller.sh` will produce
Caller 1 /Users/corti/tmp
Callee 1 /Users/corti/tmp
Callee 2 /tmp
Caller 2 /Users/corti/tmp
As you see when printing `Caller 2` the parent process has still the same working directory