How to construct the Lie group homomorphism $SU(2) \to SO(3)$ via the Lie algebra isomorphism $\Lambda:{\frak su}(2)\to{\frak so}(3)$? The Lie algebra of $ \mathfrak{so(3)} $ and $ \mathfrak{su(2)} $ are respectively 
$$ [L_i,L_j] = i\epsilon_{ij}^{\;\;k}L_k $$
$$ [\frac{\sigma_i}{2},\frac{\sigma_j}{2}] = i\epsilon_{ij}^{\;\;k}\frac{\sigma_k}{2} $$
And of course, there is an isomorphism between these two algebras, 
$$ \Lambda : \mathfrak{su(2)} \rightarrow \mathfrak{so(3)} $$ 
such that $ \Lambda(\sigma_i/2) =L_i  $
Now is it possible, using $\Lambda$, to construct a group homomorphism between $SU(2)$  and $SO(3)$?
I was checking up on Lie group homomorphism, and in Wikipedia, there is a beautiful image

In this image's language, how are $\phi$ and $\phi_*$ related to each other (just like the algebra and group elements are).
Note : I know there is a one-to-two homomorphism between these two groups which can be directly found using the group elements. I am not looking for this.
EDIT 1 : In $ SL(2,\mathbb{R}) $ the generators, say $X_1,X_2,X_3$, they obey the following commutation rules : 
$$ [X_1,X_2] = 2X_2 $$
$$ [X_1,X_3] = -2X_3 $$
$$ [X_2,X_3] = X_1 $$
And in the case of $ SO(3) $ with a different basis, $ L_{\pm} = L_1 \pm i L_2 $ and $ L_z = L_3  $ with the commutators being,
$$ [L_z,L_{\pm}]= \pm L_{\pm} $$
$$ [L_+,L_-]= 2 L_z $$
This algebra is very similar to the algebra of the previous one, so why is that we can't define a map ?
EDIT 2:
Can the group homomorphism between these two groups be written like this (Something like what I expected) :
$$ R = \exp(\sum_k i t_k L_k) = \exp\left(\sum_k i t_k \frac{\sigma_k}{2}\right) = \exp\left(\sum_k i t_k \frac{1}{2}ln(U_k)\right) $$
Now this seems like the map $\phi$,
$$ R = \phi(U) = \exp\bigg(\sum_k i t_k \frac{1}{2}ln(U_k)\bigg) $$
 A: So I take it you are clearly aware that the big A Adjoint representation is the homomorphism you're after in this case, so you're seeking a more general method.
Also, I'm assuming you know that the homomorphism of Lie algebras can only lift to a group homomorphism if the homomorphism's domain is simply connected, in which case there is a unique group homomorphism with the given algebra homomorphism as its Lie map. In this case, we're in the clear because $SU(2)$ is simply connected. Page 73 through 76 of:
Anthony Knapp, "Lie Groups Beyond an Introduction" 
can then help you. Knapp gives you two methods of systematically constructing the simply connected Lie group: the first leaves you with differential equations for the left / right invariant vector fields, the second I believe is the same as V. Moretti's Answer. 
A final "method" is to used Ado's theorem, which assures us that we can always realise a Lie algebra as a matrix Lie algebra; there is even an explicit software algorithm for this:
W. A. De Graaf, "Constructing Faithful Matrix Representations of Lie Algebras"
but if you can understand this algorithm, you are doing better than I (this paper has so far defeated me). Once you have a matrix algebra, you can used the matrix exponential to construct a neighbourhood of the identity, indeed the whole group if the latter is compact; as in V. Moretti's Answer the Lie algebra does not exponentiate to the whole group for noncompact groups (as far as I am aware, the problem of exactly what in a noncompact Lie group can be realised as an exponential of a Lie algebra element is to some extent still an open problem). 
So, once you have the Lie group, you can in principle construct the universal cover with homotopy classes and carve out the discrete centre $\mathcal{Z}_d$ of the universal cover. Your original group will have as its fundamental group the quotient group of $\mathcal{Z}_d$ and one of its (normal) subgroups.
A: First notice that the generators are $-i\sigma_k/2$ and $-iL_k$, since the groups are real Lie groups and thus the structure tensor must be real. 
The answer to your question is positive. In principle it is enough to take the  exponential of the Lie algebra isomorphism and a surjective Lie group homomorphism arises this way
  $\phi : SU(2)\to SO(3)$: $$\phi\left(\exp\left\{-\sum_k t^k i\sigma_k/2\right\}\right) =\exp\left\{-\sum_k t^k iL_k\right\}\:.$$
The point is that one should be sure that the argument in the left-hand side covers the whole group. For the considered case, this is true because $SU(2)$ is compact.
If you instead consider no compact Lie groups, like $SL(2,\mathbb C)$, the exponential does not cover the group. However it is possible to prove that products of exponential do. In that case a product of two exponentials is sufficient, in practice decomposing an element of $SL(2,\mathbb C)$ by means of the polar decomposition, mathematically speaking, or as a (unique) product of a rotation and a boost physically speaking. 
