A map that's 1-1 but not onto I've got some confusion about the definition of a 1-1 map. 
When I searched  for "1-1 correspondence" on Wikipedia, I  got redirected to the "bijection" page. So I think the two words mean just the same thing. 
Wikipedia definition of a bijection is as follows:

For a pairing between X and Y (where Y need not be different from X) to be a bijection, four properties must hold:
  each element of X must be paired with at least one element of Y,
  no element of X may be paired with more than one element of Y,
  each element of Y must be paired with at least one element of X, and
  no element of Y may be paired with more than one element of X.

Property 3 says that Y (codomain) must be contained in the range of this map, and hence the map is surjective, or onto. 
However, I've read that in linear algebra if a vector space $V$ is infinite-dimensional, there might exist a linear map $B\in\mathcal L(V)$ such that $B$ is 1-1 but not onto. 
I don't understand how, since I think 1-1 must imply onto, as indicated by the definition. 
Could you point out where I'm wrong and give me one example of such a map? Best regards!
 A: The term "one-to-one" is ambiguous. Some authors use it to mean "injective", while others use it to mean "bijective" (i.e. both injective and onto). "injective" means properties 1, 2, and 4 hold. (Note that properties 1 and 2 are just the definition of a function, so an "injective function" usually is defined as a function with property 4.
The term "injective" is never ambiguous, nor are the terms "onto", "surjective" (which means the same as "onto"), or "bijective." In the context of linear maps (or maps of groups, rings, etc.), you may hear "isomorphism" instead of "bijective map" since a bijective linear map (or group homomorphism, etc.) is an isomorphism (which is defined as a map having a linear inverse).
I'd avoid using "one-to-one" whenever possible.
Wikipedia is using "one-to-one" to mean "bijective", while the linear algebra book you read is using it to mean "injective".
It's a theorem in linear algebra that if $V$ is finite dimensional and $L: V \rightarrow V$ is a linear map, then it is injective iff it is onto; thus, in this case, the two meanings coincide. (Note, however, that the map $L: \mathbb{R} \rightarrow \mathbb{R}^2$ given by $x \mapsto (x,0)$ is injective but not onto!), so the author can be forgiven for using confusing terminology.
Also, note that for vector spaces (or groups, etc), being injective is the same thing as having a trivial kernel: $f(x) = f(y)$ iff $f(x - y) = 0$, so if no non-trivial elements map to $0$, then no pair of distinct elements map to the same thing.
For a counterexample in the infinite dimensional case, consider the vector space $V$ consisting of infinite sequences of real numbers $(a_1, a_2, \ldots, a_n, \ldots)$ with component-wise addition and scalar multiplication. (we don't require that they converges or anything). Then the map $S: V \rightarrow V$ given by $(a_1, a_2, \ldots, a_n, \ldots) \mapsto (0, a_1, a_2, \ldots)$ is linear and injective, but not onto (everything in the image has first coordinate $0$). Clearly this kind of example doesn't work in the finite dimensional case, since if we "shift" everything by one place like this, there is nowhere we can "shift" the last coordinate to.
A: In general, the terminology "1-1" (not to be confused with 1-1 correspondence) means injective and onto means surjective. If a function $f$ is injective and surjective, it is said to be bijective, or sometimes 1-1 correspondence. 
There are many examples which $f$ is injective but not surjective. In particular, if you look for one on vector spaces, consider the right-shift operator
$$T: \mathit{l}^2 \rightarrow \mathit{l}^2, \qquad T(x) = (0, x_1, x_2, x_3, \ldots) \text{ for } x = (x_1, x_2, x_3, \ldots) \in \mathit{l}^2$$
It should be easy to verify $T$ is injective but not surjective. If you want an example which is surjective but not injective, you can consider the left-shift operator on $\mathit{l}^2$.
In the context of vector spaces, if $V$ and $W$ are of equal finite dimension, then a linear operator $T:V \rightarrow W$ is injective if and only if $T$ is surjective. This is very special for finite dimensional vector spaces, as even for infinite dimensional vector spaces this property fails (just take the two-shift operators with $V=W=\mathit{l}^2$ as example).
