How does "arch -arch" work to enable Rosetta and force binary architecture?

Question

When using Rosetta on macOS running on an ARM processor, the arch command can be used to force a wrapped command to execute as a specific architecture. In other words, arch -arch x86_64 mycommand will try to launch mycommand as an x86/Intel program. If mycommand is ARM-only, it will fail; if mycommand is a universal binary, it will be run as x86.

But how does the arch -arch ... command actually work? Specifically, I have these questions:

• By what mechanism does arch -arch $arch $program select a $arch from a universal binary $program and run it?
• By what mechanism does arch -arch $arch $program try to run a non-universal binary $program as the selected architecture $arch, causing a failure if the binary doesn't support it?
• Is it possible for another program to do what arch does here, without shelling out to arch? Even if it's possible-but-not-recommended/a private macOS API, I'm still interested in how it happens.

I'm asking this out of curiosity, not because I need to get around what arch does.

I've tried:

• Looking into the guts of lipo while making universal binaries. lipo seems to be concatenating binaries for different architectures together in a Mach-O "fat binary" container with some magic bits at the top to indicate what contents are where. This helps a little bit with the first question above, but doesn't help with the second or third.
• Looking into fat binary parsing tools like mach_object. It seems theoretically possible for me to extract the appropriate sub-binary from a fat binary, dump it to a temp file, and execute it, but that seems like it would have all sorts of problems (CWD would be wrong, writing a file just to launch a program seems wasteful).