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let $D$ be a smooth manifold with boundary. Is it true that the Euler characteristic of $D$ is equal to 1 if and only if $D$ is simply connected? For domains in $\mathbb{R}^2$ this should be true, but I am not sure for general manifolds.

Edit: The answer below explains that this is false in general. But it is true for 2-dimensional manifolds with non-empty boundary.

Does anyone know a good source which explains such characteristics of the Euler characteristic?

Best wishes

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    Please do not change the question in a way which invalidates an already posted answer. – 2017-01-07
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    Sorry for that. I hope it is now better. – 2017-01-07

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This is false in general. Removing $k$ open balls from $S^3$ leaves a manifold of Euler characteristic $-k$, and that manifold is simply connected because it is homotopy equivalent to removing $k-1$ points from $\mathbb{R}^3$.

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    Thank you. Is it maybe true for 2 dimensional manifolds with boundary? – 2017-01-07
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    Yes, if you are assuming that the boundary is nonempty. Otherwise the projective plane is a counterexample. – 2017-01-07
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    Ok thank you very much for this clarification. Do you know a reference where I can find a proof? – 2017-01-07
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    This is an application of the classification of surfaces. – 2017-01-07