Isometries of $\mathbb{R}^n$ Let $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$ be such that $\left\| f(x)-f(y)\right\| =\left\| x-y\right\|$.  Is $f$ necessarily surjective?
If this is so, you can prove (Mazur-Ulam Theorem) that $f$ is affine, and hence you could classify all isometries of $\mathbb{R}^n$.  However, at the moment, I can't think of any good ideas to prove that $f$ is surjective.  For that matter, is it even the case that $f$ must be surjective?
Any ideas would be most welcomed.
Thanks much!
 A: Let $ f $ be an isometry of $ \mathbb{R}^n $ ( i.e. a map $ f : \mathbb{R}^n \rightarrow \mathbb{R}^n $ such that $ || f(x) - f(y) || = || x - y || $ for all $ x,y $ ). We can write $ f(x) = f(0) + ( f(x) - f(0) ) $ and now $ g(x) := f(x) - f(0) $ is an isometry fixing $ 0 $ ( especially $ g $ preserves norm, i.e. $ || g(x) || = || x || $ for all $ x $ ).  We'll now focus on $ g $.
Notice $ g $ preserves inner product i.e. $ \langle g(x), g(y) \rangle = \langle x, y \rangle $ for all $ x, y $ ( since expanding $ \langle g(x) - g(y), g(x) - g(y) \rangle = \langle x - y, x - y \rangle $ gives $ || g(x) ||^2 - 2 \langle g(x), g(y) \rangle + || g(y) ||^2 = || x ||^2 + || y ||^2 - 2 \langle x, y \rangle $ and so $ \langle g(x), g(y) \rangle = \langle x, y \rangle $ ).
Turns out $ g $ is linear too : Expanding $ || g(x+y) - g(x) - g(y) ||^2 $ as $  \langle g(x+y) - g(x) - g(y), g(x+y) - g(x) - g(y) \rangle $ gives \begin{align*}
&|| g(x+y) ||^2 + || g(x) ||^2 + || g(y) ||^2 - 2 \langle g(x+y), g(x) \rangle - 2 \langle g(x+y), g(y) \rangle + 2 \langle g(x), g(y) \rangle \\
&= || x + y ||^2 + ||x||^2 + || y ||^2 - 2 \langle x + y, x \rangle - 2 \langle x + y, y \rangle + 2 \langle x, y \rangle\\
&= 0.
\end{align*}
Therefore $ g(x+y) = g(x) + g(y) $ for all $ x,y $. Similarly we can show $ g(ax) = ag(x) $ for all $ a \in \mathbb{R} $ and $ x \in \mathbb{R}^n $.
So $ g(x) = Ax $ for some $ n \times n $ matrix $ A $. Notice $ g(e_j) = A_j $, where $ (e_1, \cdots, e_n) $ is the standard basis of $ \mathbb{R}^n $ and $ A_1, \cdots, A_n $ are the columns of $ A $. Since $ \langle g(e_i), g(e_j) \rangle = \langle e_i, e_j \rangle $, we see $ \langle A_i, A_j \rangle $ is $ 0 $ if $ i \neq j $ and $ 1 $ if $ i = j $. Therefore $ A^T A = I $, i.e. $ A $ is an orthogonal matrix.
To summarise : Let $ f $ be an isometry of $ \mathbb{R}^n $. Then $ f(x) = f(0) + Ax $ for some orthogonal matrix $ A $.
( So especially the $ f $ here is bijective, answering the original question )
Edit : Treil's "Linear Algebra done Wrong" mentioned in the comments seems to take the same approach ( which is also outlined in above answers ), making this quite redundant. Nevertheless I'll leave it undeleted.
A: First assume $f$ fixes the origin. Show that $f$ preserves the inner product. Then you can show that $f$ is linear. Since you can translate a general isometry $f$ to obtain a new isometry that fixes the origin, $f$ must be surjective, and in fact given by $Ax+b$, where $A$ is an orthogonal matrix and $b$ is a vector.
A: Let's assume WLOG that $f(0) = 0$. For every $r$, it follows that $f$ defines an isometry from the sphere of radius $r$ to the sphere of radius $r$. 
Proposition: Any isometry $f : X \to X$ of a compact metric space is bijective.
Proof. $f$ is clearly injective. Suppose $f$ is not bijective. Then $f(X)$ is compact, so given $x \in X \setminus f(X)$ the distance $\text{dist}(x, f(X))$ is positive. Pick $\epsilon < \text{dist}(x, f(X))$. Let $N$ be the smallest positive integer for which $X$ admits a cover by $N$ open sets of diameter less than $\epsilon$. No such set containing $x$ can intersect $f(X)$, but by pulling back along $f$ it follows that we can find a cover of $X$ by $N-1$ open sets of diameter less than $\epsilon$; contradiction.
(In fact any isometry of a compact metric space is a homeomorphism, since a continuous bijection from a compact space to a Hausdorff space is necessarily closed.)
Apparently there are counterexamples to the above when $X$ is not compact, but I don't know any nice ones off the top of my head. 
A: Jonathan,
Yeah I thought that too at first, but you can show an isometry of $\mathbb{R}^n$ fixing the origin is linear without assuming that it's surjective. The key is the inner product, which, of course, you don't have in a general normed vector space. Preserving the inner product and fixing the origin implies that the map is linear (a great exercise). Then since it's injective and we're in finite and equal dimensions, it's also surjective. I found this thread because I had the exact same question as you. There are geometric proofs of surjectivity in $\mathbb{R}^2$ involving triangles or circles as well.
