I am reading "Topology 2nd Edition" by James R. Munkres.
The following exercise is in this book:
Exercise 1(a) on p.44 in section 6:
Make a list of all the injective maps $$f:\{1,2,3\}\to\{1,2,3,4\}.$$ Show that none is bijective. (This constitutes a direct proof that a set $A$ of cardinality three does not have cardinality four.)
I made a list of all the injective maps $f:\{1,2,3\}\to\{1,2,3,4\}.$
I firmly believe that the following list is correct.
But how can I prove the following list is a list of all the injective maps $f:\{1,2,3\}\to\{1,2,3,4\}$ rigorously?
$f_{1}:\{1,2,3\}\to\{1,2,3,4\}$ such that $f_{1}(1)=1, f_{1}(2)=2, f_{1}(3)=3.$
$f_{2}:\{1,2,3\}\to\{1,2,3,4\}$ such that $f_{2}(1)=1, f_{2}(2)=2, f_{2}(3)=4.$
$f_{3}:\{1,2,3\}\to\{1,2,3,4\}$ such that $f_{3}(1)=1, f_{3}(2)=3, f_{3}(3)=2.$
$f_{4}:\{1,2,3\}\to\{1,2,3,4\}$ such that $f_{4}(1)=1, f_{4}(2)=3, f_{4}(3)=4.$
$f_{5}:\{1,2,3\}\to\{1,2,3,4\}$ such that $f_{5}(1)=1, f_{5}(2)=4, f_{5}(3)=2.$
$f_{6}:\{1,2,3\}\to\{1,2,3,4\}$ such that $f_{6}(1)=1, f_{6}(2)=4, f_{6}(3)=3.$
$f_{7}:\{1,2,3\}\to\{1,2,3,4\}$ such that $f_{7}(1)=2, f_{7}(2)=1, f_{7}(3)=3.$
$f_{8}:\{1,2,3\}\to\{1,2,3,4\}$ such that $f_{8}(1)=2, f_{8}(2)=1, f_{8}(3)=4.$
$f_{9}:\{1,2,3\}\to\{1,2,3,4\}$ such that $f_{9}(1)=2, f_{9}(2)=3, f_{9}(3)=1.$
$f_{10}:\{1,2,3\}\to\{1,2,3,4\}$ such that $f_{10}(1)=2, f_{10}(2)=3, f_{10}(3)=4.$
$f_{11}:\{1,2,3\}\to\{1,2,3,4\}$ such that $f_{11}(1)=2, f_{11}(2)=4, f_{11}(3)=1.$
$f_{12}:\{1,2,3\}\to\{1,2,3,4\}$ such that $f_{12}(1)=2, f_{12}(2)=4, f_{12}(3)=3.$
$f_{13}:\{1,2,3\}\to\{1,2,3,4\}$ such that $f_{13}(1)=3, f_{13}(2)=1, f_{13}(3)=2.$
$f_{14}:\{1,2,3\}\to\{1,2,3,4\}$ such that $f_{14}(1)=3, f_{14}(2)=1, f_{14}(3)=4.$
$f_{15}:\{1,2,3\}\to\{1,2,3,4\}$ such that $f_{15}(1)=3, f_{15}(2)=2, f_{15}(3)=1.$
$f_{16}:\{1,2,3\}\to\{1,2,3,4\}$ such that $f_{16}(1)=3, f_{16}(2)=2, f_{16}(3)=4.$
$f_{17}:\{1,2,3\}\to\{1,2,3,4\}$ such that $f_{17}(1)=3, f_{17}(2)=4, f_{17}(3)=1.$
$f_{18}:\{1,2,3\}\to\{1,2,3,4\}$ such that $f_{18}(1)=3, f_{18}(2)=4, f_{18}(3)=2.$
$f_{19}:\{1,2,3\}\to\{1,2,3,4\}$ such that $f_{19}(1)=4, f_{19}(2)=1, f_{19}(3)=2.$
$f_{20}:\{1,2,3\}\to\{1,2,3,4\}$ such that $f_{20}(1)=4, f_{20}(2)=1, f_{20}(3)=3.$
$f_{21}:\{1,2,3\}\to\{1,2,3,4\}$ such that $f_{21}(1)=4, f_{21}(2)=2, f_{21}(3)=1.$
$f_{22}:\{1,2,3\}\to\{1,2,3,4\}$ such that $f_{22}(1)=4, f_{22}(2)=2, f_{22}(3)=3.$
$f_{23}:\{1,2,3\}\to\{1,2,3,4\}$ such that $f_{23}(1)=4, f_{23}(2)=3, f_{23}(3)=1.$
$f_{24}:\{1,2,3\}\to\{1,2,3,4\}$ such that $f_{24}(1)=4, f_{24}(2)=3, f_{24}(3)=2.$
In order to have an injective map from $\{1,2,3\}$ into $\{1,2,3,4\}$, the image should have $3$ elements. We have four subsets of cardinality 3 and each one of those sets has $3!$ permutations.
That makes $4(3!)=24$ possible sets of images.
Thus there are $24$ injective functions from $\{1,2,3\}$ into $\{1,2,3,4\}$