Take tautological bundle over $\mathbb{C}P^n$ less the zero section. Is this homeomorphic to $\mathbb{C}^{n+1}-{0}$? It appears that (non zero vector in $\mathbb{C}^{n+1}$)$\mapsto$(line containing the vector, vector in this line) is a homeomorphism from  $\mathbb{C}^{n+1}$ to the tautological bundle over $\mathbb{C}P^n$ less the zero section, but I get very tangled in the subtleties of projective space. Could someone please confirm this/tell me I am wrong?
 A: 1) Let $E$ be the total space of the the tautological vector bundle $\mathcal O(-1)$ over $\mathbb P^n(\mathbb C)$.
It consists of all pairs $(\mathbb Cv,v)$ with $v$ a non-zero vector $v\in \mathbb C^{n+1}\setminus \{0\}$.
The sero section $Z\subset E$  of this line bundle consists of the pairs $(\mathbb Cv,0)$ [with again $v\in \mathbb C^{n+1}\setminus \{0\}$].
The regular morphism $$f:\mathbb C^{n+1}\setminus \{0\}\stackrel {\cong}{\longrightarrow}E\setminus Z: v\mapsto (\mathbb Cv,v)$$ has as inverse $$f^{-1}:  E\setminus Z         \stackrel {\cong}{\longrightarrow}\mathbb C^{n+1}\setminus \{0\}:(\mathbb Cv,v)\mapsto  v$$ and is an isomorphism of algebraic varieties.
It is then a fortiori an isomorphism of complex manifolds, a diffeomorphism of differential manifolds and, as you wished, a homeomorphism of topological spaces. 
2) If you know some elementary algebraic geometry you will find the following quite geometric:
Blowing-up  $\mathbb C^{n+1}$ at $0$ you obtain $p:\widehat {\mathbb C^{n+1}}=E\to \mathbb C^{n+1}$ with $Z=p^{-1}(0)\subset E$ as the blown-up origin.
Removing $Z$ from $E$ you obtain the restricted morphism $$ p\vert _{E\setminus Z}:E\setminus Z\stackrel {\cong}{\longrightarrow}\mathbb C^{n+1}\setminus \{0\}$$ which is an isomorphism and is none other than the morphism  $f^{-1}$ of part 1) of this answer.
A: Yes.  In fact, the map you cite,
$$\begin{align*}
\text{Tautological Line Bundle } - \text{Zero Section} & \to \mathbb{C}^{n+1} - 0 \\
(z, \ell) & \mapsto z,
\end{align*}$$
is a biholomorphism.
