Why is the image of a proper patch an open set?

Question

O'Neill's Elementary DIfferential Geometry mentions that the image of a proper patch is always a surface. This seems to be based on the proposition that the image of a proper patch is an open set. Why is this true? I see that the domain of a patch is an open set, and the mapping is continuous. Would it be true to say that the image, under a continuous mapping, of an open set is always open?

Answer 1

O'Neill's Elementary DIfferential Geometry mentions that the image of a proper patch is always a surface. This seems to be based on the proposition that the image of a proper patch is an open set. Why is this true?

This is true, because every open subset of surface is surface aswell. See every open subset of a surface is a surface.

Would it be true to say that the image, under a continuous mapping, of an open set is always open?

Well no. Consider constant function $\mathbb R^2 \rightarrow \mathbb R^2$ with standard metric. Image is a single point which is closed.

The keyword is here proper. A proper function takes open sets and maps them to open sets.