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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?

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    Wikipedia is right. Do you have a precise statement of what your lecturer said? – 2017-01-08
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    @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. – 2017-01-08
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    Or even here: https://en.wikipedia.org/wiki/Isometry section: Formal definitions. it seems that they attribute "bijective" to isometry, so there can be a non-bijective? – 2017-01-08
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    Well here, http://mathworld.wolfram.com/Isometry.html they define it as bijective... really confusing.. – 2017-01-08
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    Your link https://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. – 2017-01-08
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    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](http://math.stackexchange.com/questions/116195). Without completeness of $M$, you have examples such as $x \mapsto x + 1$ on the set of positive reals. – 2017-01-08
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    @AndrewD.Hwang: You need simply connected target not just connected. – 2017-01-09
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    @MoisheCohen: Whoops...yes, thank you. – 2017-01-09

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