 CRC32(32-bit buffer) is a bijective function

I've always been interested if this is true or not?

#define POLY 0xedb88320

#include <inttypes.h>
#include <stdio.h>
#include <string.h>

// copypasted from https://rosettacode.org/wiki/CRC-32#C
uint32_t
rc_crc32(uint32_t crc, const char *buf, size_t len)
{
static uint32_t table;
static int have_table = 0;
uint32_t rem;
uint8_t octet;
int i, j;
const char *p, *q;

/* This check is not thread safe; there is no mutex. */
if (have_table == 0) {
/* Calculate CRC table. */
for (i = 0; i < 256; i++) {
rem = i;  /* remainder from polynomial division */
for (j = 0; j < 8; j++) {
if (rem & 1) {
rem >>= 1;
rem ^= POLY;
} else
rem >>= 1;
}
table[i] = rem;
}
have_table = 1;
}

crc = ~crc;
q = buf + len;
for (p = buf; p < q; p++) {
octet = *p;  /* Cast to unsigned octet. */
crc = (crc >> 8) ^ table[(crc & 0xff) ^ octet];
}
return ~crc;
}

main()
{
uint32_t tmp=0;

while (1)
{
printf("%08x\n", rc_crc32(0, &tmp, sizeof(uint32_t)));
if (tmp==0xffffffff)
break;
tmp++;
};

return 0;
}

This program generates all possible $2^{32}-1$ values. I tried these known CRC-32 polynomials: 0xedb88320, 0x82F63B78, 0xEB31D82E, 0x80000000.

I tried even random polynomials, all works: 0xffffffff, 0x82345678.

So this is a a perfect hash function, with no collisions.

But this polynomial is faulty: 0x12345678.

Here is a simple explanation. If you divide all possible 32-bit numbers by a 32-bit constant, you'll get remainders between 0 and this 32-bit constant.

CRC is a calculation of remainder over GF(2): input message is a dividend, polynomial is a divisor, CRC output is a remainder. So no wonder that outputs are all possible 32-bit values if a polynomial has highest bit set (0x80000000).

This would generalize to all CRC polynomials of other widths.

The 'Mathematical recipes' book has a section devoted to explanation of CRC algorithm as a division. 