Surjectivity of Hurewicz fibrations Proposition 4H.1 in Hatcher's Algebraic Topology states that any cofibration is injective. Since cofibrations and fibrations are dual concepts (in a way that I can't yet state formally, though), this makes me think fibrations are always surjective. However, I was unable to prove this. In fact, the map from an empty space is a non-surjective fibration. Are there non-trivial examples of such fibrations, or are they in fact surjective?
 A: Here are two useful facts.
(1) If $p : E \to B$ is a fibration, then also the surjectve map $\tilde{p} : E \stackrel{p}{\to} p(E)$ is a fibration.
This follows immediately from the definition.
(2) If $p : E \to B$ is a fibration, then $p(E)$ is a union of path components of $B$.
Let $x \in p(E)$ and let $P$ denote the path component of $x$ in $B$. For $y \in P$ choose a path $u : I \to B$ such that $u(0) =x, u(1) = y$. Let $X = \{ \ast \}$ be a one-point space. Define $f : X \to E, f(\ast) = e$ with an arbitrary $e \in p^{-1}(x)$ and $F : X \times I \to B, F(\ast,t) = u(t)$. $F$ has a lift $\tilde{F} : X \times I \to E$. Then $\tilde{u} : I \to E, \tilde{u}(t) = \tilde{F}(\ast,t)$ is a lift of $u$. In other words, each path starting in $p(E)$ can be lifted to $E$ ("path lifting property"). This means that $u = p \circ \tilde{u}$ is a path in $p(E)$ and we get $y \in p(E)$. Therefore $P \subset p(E)$.
It follows from (2) that $C = B \backslash p(E)$ is also a union of path components of $B$. In general $C$ it is non-empty, but it is separated from $p(E)$ in the sense that there are no paths connecting $C$ and $p(E)$.
Therefore, if you start with a surjective fibration $p : E \to B$, you can enlarge $B$ to any $B'$ such that $B' \backslash B$ is a union of path components of $B'$ and get again a fibration $p' : E \to B'$. We may regard this as a sort of "trivial base extension" which does not bring something new. In other words, it suffices to focus attention to surjective fibrations. 
A: Fibrations need not be surjective. For example, let $f : E \to B$ be a fibration and let $i : B \to B \sqcup *$ be the inclusion, then $i\circ f : E \to B\sqcup *$ is again a fibration, but it is not surjective, even if $f$ is.
