Injective self-map - check my proof Let $f:[N]\to [N]$ be an injective self-map where $[N]:= \{1, 2, ..., N\}$ is a finite set of $\mathbb{N}$. Show that $f$ is bijective.
My proof:
As $f$ is injective and finite we simply count all images of $N$ which are $N$-many. As $f([N]) \subseteq [N] $, we conclude $f([N]) = [N] $ as both sets have the same cardinality. So $f$ is surjective and therefore bijective. $q.e.d.$
Is this correct?
The approach of our professor is different and I don't understand one step he did:
He took the union of all preimages $\bigcup_{k=1}^N \{f^{-1}(k)\}$, which are also $N$-many and uses injectivity to argue that all $\{f^{-1}(k)\}$ are pairwise disjoint and therefore $\bigcup_{k=1}^N \{f^{-1}(k)\} = [N]$. 
Why holds $\bigcup_{k=1}^N \{f^{-1}(k)\} = [N]$? 
$\bigcup_{k=1}^N \{f^{-1}(k)\} \subseteq [N]$ is certainly true but I don't understand the other subset-property $\supseteq$ which finally shows that both sets are euqal?
Can someone explain this step to me?
 A: Suppose $\bigcup_{k=1}^N f^{-1}(k) \neq [N]$. So there exists $n \in [N]$ such that $n \notin f^{-1}(k)$ for any $k=1,\dots,N$. In other words, $f(k) \neq n$ for any $k=1,\dots,N$. By the pigeonhole principle, since there are $N$ values of $k$ and at most $N-1$ possible values of $f(k)$, there must exist $k_1 \neq k_2$ such that $f(k_1)=f(k_2)$. This contradicts the fact that $f$ is injective.
A: I'm going to write $[n]$ for your $[N]$.
Your proof is good, just add after "as both sets have the same cardinality" the following "and $[n]$ is the only subset of $[n]$ with cardinality $[n]$".
It sounds dumb but for sets of higher cardinalites, it may be false, i.e. $f[X]\subseteq X$ with $\lvert f[X]\rvert=\lvert X\rvert$, but $f[X]\ne X$. For example, take $X=\mathbb N$ and the map $n\mapsto n+1$.
What your professor did is essentially the same. You used that injective implies that the image of any $k\in[n]$ is singleton, then unioning $[n]$-many singleton sets gives a set of cardinality $[n]$. Your professor used that injective implies that the preimage of any $k\in[n]$ is singleton, . . .  and so on.

Here's an alternative proof that I wrote earlier.
