# Is an isometry a diffeomorphism?

It seems that isometry does not have to be bijective (surjective) as I learnt from my lecture. But why on https://en.wikipedia.org/wiki/Isometry_(Riemannian_geometry) defined isometry as diffeomorphism? So it is bijective? A bit confusing.

Which is right?

• Wikipedia is right. Do you have a precise statement of what your lecturer said? Jan 8, 2017 at 19:22
• @gj255 The embedding $P:\mathbb{R}^2\rightarrow\mathbb{R}^3$ with $P(x) = (x_1,x_2,0)$ is an isometric embedding. It is not surjective. Maybe, one strictly distinguishes between an isometry and an isometric embedding. This could be the source of confusion. Jan 8, 2017 at 19:46
• Or even here: en.wikipedia.org/wiki/Isometry section: Formal definitions. it seems that they attribute "bijective" to isometry, so there can be a non-bijective?
– Upc
Jan 8, 2017 at 19:58
• Your link en.wikipedia.org/wiki/Isometry is informative with respect to your question. There they say "usually assumed to be bijective". This already implies that there are different versions in use. I recommend sticking to the version of your teacher and keeping in mind that there might be a different version in literature. Jan 8, 2017 at 20:50
• If $M$ is a complete, connected Riemannian manifold and $N$ is connected, then a local isometry $f:M \to N$ is bijective: How to go from local to global isometry. Without completeness of $M$, you have examples such as $x \mapsto x + 1$ on the set of positive reals. Jan 8, 2017 at 22:56