Surjectivity of isometries I have read different posts about this subject, all focused on very specific assumptions (compactness, in $\mathbb{R}^N$, etc.). My question aims at a unifying goal.
Let $(X,d_X)$ and $(Y,d_Y)$ metric spaces and $\sigma:X\to Y$ a distance preserving map (isometry). Is it true that $\sigma$ is surjective if (maybe, and only if)  $$\mathrm{diam}_X(X)\geq \mathrm{diam}_Y(Y) \, .$$
Update. [the answers below are illuminating]
Add the condition on $\sigma$ that there exists $x\in X$ such that balls in $X$ centered at $x$ are sent by $\sigma$ to balls in $Y$ centered at $\sigma(x)$.
Update 2. Even this condition is not sufficient as the example (of @Dry Bones) in the comments show. Any new guess on how to fix the hypotheses is welcome!
 A: No, even if you are working in Hilbert space. Let $H$ be a separable Hilbert space. Fix an orthonormal base $\{e_i\mid i\in\mathbb{N}\}$. Define a unilateral shift operator $U:H\rightarrow H$ by $U(e_i)=e_{i+1}$ (of course, extended by linearity and continuity), then $U$ is an isometry but not surjective.
A: Complementarily to Danny Pak-Keung Chan's answer, even if the diameters are finite, it doesn't work. For example, the inclusion of a one-dimensional interval in a two-dimensional disk with the usual metrics induced from the plane is an isometry, satisfies the prescribed inequality, but isn't surjective at all.
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Here's another example that must be dealt with in order to find a valid version of your statement.
$X$ is the blue 'ellipse', $x$ is the point at the center of $X$, and $Y=X\cup\{p\}$, where $p$ is the isolated blue point on the right. The isometry is the inclusion $X\rightarrow Y$. All the aforementioned conditions hold, even the stronger '$\sigma$ takes closed balls centered at $x$ to closed balls centered at $\sigma(x)$', but the isometry isn't surjective.
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