When working with topological vector spaces (say $X,Y$), the term “embedding” is often used for a continuous injection $f:X\rightarrow Y$.
Now, $f$
is of course a bijection onto its image, but it's not necessarily a homeomorphism (as we would normally require of an embedding in topology).
Can anyone explain to me exactly what we're loosing by not having $X$
and $f(X)$ homeomorphic? And is there something special about linear spaces that makes whatever we're missing (more or less) irrelevant?
I suppose that the subspace topology on $f(X)$
must in some sense be coarser than the topology on $X$
if the inverse of $f$
(defined on the range of $f$
) fails to be continuous
Best Answer
Think of the torus as the quotient of $\mathbb R^2$ in the usual way: $(x, y) \mapsto (x \bmod 1, y \bmod 1)$.
Now look at the line $y = \pi x$ in the plane. Under this quotient, it becomes a curve in the torus, and the map is 1-1 and continuous, but locally, in the image, it doesn't look the way you expect, i.e., like a line in $\mathbb R^2$, where near the line there is "no other stuff".