Showing the tautological and tangent bundles of $\mathbb{CP}^1$ are non-isomorphic I'm trying to show that the tangent and tautological bundle of $\mathbb{CP}^1$ are non-isomorphic. The hint I was given was to look at the fundamental group of the complement of the zero section. For the tangent bundle, I used that the bundle has local trivializations $U_1=\mathbb{CP}^1 - [0:1]$ and $U_2=\mathbb{CP}^1 - [1:0]$. I then applied the siefert van kampen theorem
$$\pi_1 (TU_i-\sigma_0) \cong \pi_1 (U_i \times \mathbb{C}^*) \cong \pi_1 (U_i) \oplus \pi_1 (\mathbb{C}^*) \cong \mathbb{Z} a_i$$
$$\pi_1 (T(U_1 \cap U_2) - \sigma_0) \cong \pi_1 ((U_1 \cap U_2) \times \mathbb{C}^*) \cong \pi_1((U_1 \cap U_2)) \oplus \pi_1(\mathbb{C}^*) \cong \mathbb{Z}c \oplus \mathbb{Z}d$$
Then the fundamental group will be $\mathbb{Z}a_1 *\mathbb{Z}a_2$ quotiented by the images of c and d. c will be homotopic to a point and so the fundamental group will be $\mathbb{Z}a_1 *\mathbb{Z}a_2$ quotiented by the image of $\mathbb{Z}d$. Since the former has two generators and the latter only one, this cannot be trivial.
What I have so far I believe is correct, please tell me otherwise if it is not. What remains is to show that the tautological bundle minus the zero section has trivial fundamental group. I think that this is because it can be deformation retracted onto the 3-sphere but I'm not sure how to show this. Any help would be greatly appreciated.
 A: First, the easier question (I will skip most details): The tautological bundle is given by 
$$ L = \{ (\ell, v) \in \mathbb{CP}^1 \times \mathbb C^2 : v \in \ell\}.$$
Thus there is a hermitian metric defined on each fiber $\ell$ and one can consider the sphere bundle of $L$ (which is the deformation retract of $L-\sigma_0$). This is exactly $\mathbb S^3$ inside $\mathbb C^2$. Thus $L-\sigma_0$ has trivial fundamental group. 
However, the more serious problem is the following argument:

Then the fundamental group will be $\mathbb{Z}a_1 *\mathbb{Z}a_2$ quotiented by the images of c and d. c will be homotopic to a point and so the fundamental group will be $\mathbb{Z}a_1 *\mathbb{Z}a_2$ quotiented by the image of $\mathbb{Z}d$. Since the former has two generators and the latter only one, this cannot be trivial.

In the above statement, you never used the fact that you are dealing with tangent bundle. If what you said is correct, it is true for all line bundle over $\mathbb {CP}^1$. But this is not true as we see from the tautological bundle. 
So, what's wrong? Indeed this is quite hidden in the argument. You wrote

c will be homotopic to a point ...

well, is this really true? Let's think clearly how you identify this $"c"$: It is the equator $E$ in $U_1\cap U_2$, but you cannot choose $c$ to be $E \times \{0\}$ since $0\notin \mathbb C^*$. So let say you choose 
$$[c] = [E\times \{1\}] \in \pi_1(U_1\cap U_2 \times \mathbb C^*)$$
and assume that the inclusion $i_1 :T(U_1\cap U_2)\setminus\sigma_0 \to TU_1\setminus\sigma_0$ is given by 
$$ U_1\cap U_2 \times \mathbb C^* \to U_1 \times \mathbb C^*, \ \ \ (z, v)\mapsto (z,v).$$
So far so good, using this choice we have $i_1[c]=0$ since like you said, $i_1(c)$ shrinks to a point in $U_1 \times \mathbb C^*$. 
What about $i_2 : T(U_1\cap U_1)\setminus \sigma_0 \to TU_2\setminus \sigma_0$? Indeed under the trivialization, $i_2$ is given by 
$$ U_1\cap U_2 \times \mathbb C^* \to U_2 \times \mathbb C^*, (z, v) \mapsto (z^{-1}, -z^{-2} v)$$
Thus if we write $c$ as mapping: 
$$c : \mathbb S^1 \to U_1\cap U_2 \times \mathbb C^*, \ \ \ c(\theta) = (e^{i\theta}, 1),$$ 
then 
$$i_2(c) (\theta) = (e^{-i\theta}, e^{i(-2\theta+\pi)}).$$
The first component still shrink to a constant curve in $U_2$, but the second component gives rises to nontrivial loop on the fiber. Indeed, $i_2[c] = a^{-2}_2$. So in the tangent bundle case, one of the relation is given by $a^2_2 = 1$. 
It is easy to find the relation for $d$: We have $i_1[d] = a_1$ and $i_2[d] =a_2$. Thus another relation is $i_1[d] i_2^{-1} [d] = 1$, or $a_1 = a_2$. Thus $\pi_1(T\mathbb{CP}^1\setminus \sigma_0)$ is the free group generated by one element $a_2$ with $a_2^2 = 1$, so it's $\mathbb Z /2\mathbb Z$. 
One can also argue as above for the tautological line bundle to conclude that it's simply connected. 
