I have few beginner questions on Lie Algebras and their representations.
First question :
In my course about $su(2)$ representations, they do the following stuff.
First : we have a basis of $su(2)$ which is : $j_a=i \frac{\sigma_a}{2}$ with $a=1..3$ (where the $\sigma_a$ are the Pauli matrices).
We have the following commutators for the element of the basis :
$$[j_a;j_b]=-\epsilon^{abc}j_c$$
To find a representation of $su(2)$ of dimension d we need to find a map $T : su(2) \rightarrow M_d(\mathbb{C})$
This map has to ensure :
(1) T is a linear map.
(2) $$ \forall (P,Q) \in su(2) ~ T([P,Q])=[T(P),T(Q)] $$
Thus for the element of the basis, if we write $T_a=T(j_a)$, we need to find 3 matrices that verifies :
$$[J_a;J_b]=-\epsilon^{abc}J_c$$
(Because to check the property on the basis is enough).
In the course I read they do everything by trying to find matrices that satisfies these commutations relations.
But at the end, how can we be sure that we have a linear map ?
I mean : how to be sure that we effectively have $T(P+Q)=T(P)+T(Q)$ ? Do we have to see it in the opposite way : we ask to $T$ to verify this property ?
Second question :
When we have a representation of an algebra, we can have immediately of representation of the group by exponentiation :
if $T$ is a representation of $su(2)$ $exp(T)$ is a representation of $SU(2)$.
To be more specific : if $g\in G$, then $\exists X \in Lie(G) ~ / ~ g=e^X$.
And we have $exp(T(X))$ which represent the element $g$ in $M_n(\mathbb{C})$.
But as the exponential is not inversible, we could have different $X$ that verifies $g=e^X$ : $g=e^{X_1}=e^{X_2}$. So we have to be sure that $e^{T(X_1)}=e^{T(X_2)}$. How do we know that ?
About the first (EDITED!) question:
If you find such $J_1, J_2$ and $J_3$ such that $[J_a;J_b]= -\epsilon^{abc}J_c$ and define $$ T(c_1 j_1 + c_2 j_2 + c_3 j_3) := c_1 T(j_1) + c_2 T(j_2) + c_3 T(j_3) = c_1 J_1 + c_2 J_2 + c_3 J_3 \qquad \text{for all } {\bf c} = (c_1, c_2, c_3) $$ then by construction your map is linear so you don't have to check.
Answer to the second question (Added later) Let us firstly clarify what you mean by
"if $T$ is a representation of $su(2)$ $exp(T)$ is a representation of $SU(2)$"
Consider the following diagram
To me, to build a representation of $SU(2)$ from the representation $T$ of $\mathfrak{su}(2)$ would mean to build a linear and bracket preserving map $$\widetilde{T} : SU(2) \to GL_d(\mathbb{C})$$ such that $$ \forall g \in SU(2) \,\,\,\, \widetilde{T}(g) := \exp(T(X)) $$ where $X \in \mathfrak{su}(2)$ is such that $\exp(X) = g$.
Now of course due to surjectivity of the exponential map $\exp : \mathfrak{su}(2) \to SU(2)$ we know that such $X$ exists for every $g \in SU(2)$. The thing is, as you mentioned, that $\exp$ is not injective so we could have two $X_1, X_2 \in \mathfrak{su}(2)$ such that $\exp(X_1)=\exp(X_2) = g$.
So in order for the map $\widetilde{T}$ to be correctly defined it is sufficient to check that $$ \exp(T(X_1)) = \exp(T(X_2)) \quad {if} \quad \exp(X_1) = \exp(X_2).$$
Now let's first see what $X_1$ and $X_2$ could be so that $\exp(X_1) = \exp(X_2)$? Let firstly $$X = a_1 j_1 + a_2 j_2 + a_3 j_3 = {\bf a}.{\bf j} = \left( \begin{matrix} i \dfrac{a_3}{2} & \dfrac{a_2}{2}+i \dfrac{a_1}{2} \\ -\dfrac{a_2}{2}+i \dfrac{a_1}{2} & \,\,-i\dfrac{a_3}{2} \end{matrix} \right) $$ We have that $$ (*) \qquad \exp(X) = \cos{\dfrac{|{\bf a}|}{2}}E_2 - \dfrac{\sin{\dfrac{|{\bf a}|}{2}}}{\dfrac{|{\bf a}|}{2}}X. $$ where $E_2$ is the identity $2\times 2$ matrix and $|{\bf a}| = \sqrt{a_1^2+a_2^2+a_3^2}$ is the norm of ${\bf a}$. Now because $\exp(X_1) = \exp(X_1)$ is equivalent to $$ \cos{\dfrac{|{\bf a}|}{2}}E_2 - \dfrac{\sin{\dfrac{|{\bf a}|}{2}}}{\dfrac{|{\bf a}|}{2}}X_1 = \cos{\dfrac{|{\bf b}|}{2}}E_2 - \dfrac{\sin{\dfrac{|{\bf b}|}{2}}}{\dfrac{|{\bf b}|}{2}}X_2 $$ where $X_1$ is parametrized by the vector ${\bf a}$ and $X_2$ - by the vector ${\bf b}$. If you write down element by element equalities you will end of with $$\cos{\dfrac{|{\bf a}|}{2}} = \cos{\dfrac{|{\bf b}|}{2}}$$ because these are the real parts of the $(1,1)$ elements of the two matrices. From here we have that $$\dfrac{|{\bf a}|}{2} = \pm\dfrac{|{\bf b}|}{2} + 2k\pi$$, or, $|{\bf a}| = \pm |{\bf b}| + 4k\pi$ for $k\in \mathbb{Z}$. Of course here $k$ should be such that both norms are positive. From here the sines are equal and this directly leads to (again by direct element-wise comparison of the matrix elements) that the vectors ${\bf a}$ and ${\bf b}$ are proportional and we can write them as $$ {\bf a} = |a|(n_1, n_2, n_3), \quad {\bf b} = |b|(n_1, n_2, n_3), \qquad n_1^2+n_2^2+n_3^2 = 1, \quad |{\bf a}| = \pm |{\bf b}| + 4k\pi. $$
Let ${\bf n} = (n_1, n_2, n_3)$. So as you can see there are actually infinitely many $X \in \mathfrak{su}(2)$ such that $\exp(X) = g$ for a fixed $g \in SU(2)$. Now let's investigate what happens to $\exp(T(X_1)) = \exp(T(X_2))$, which is equivalent to $\exp(T(X_1-X_2)) = E_2$ since $T$ is linear.
For $X_1 - X_2$ we have that it is represented by the vector ${\bf a}-{\bf b}$ which is $$ {\bf a}-{\bf b} = 2k\pi(2{\bf n}) \qquad or \qquad {\bf a}-{\bf b} = (-|{\bf b}| + 2k\pi)(2{\bf n}). $$ So because $T(n_1 j_1 + n_2 j_2 + n_3 j_3) = n_1 J_1 + n_2 J_2 + n_3 J3$ we would have $$ \exp(T(X_1-X_2)) = \left\{ \begin{array}{l} \exp{\big(2k\pi(2n_1J_1 + 2n_2J_2+2n_3J_3)\big)} = \left(\exp{\big(2n_1J_1 + 2n_2J_2+2n_3J_3\big)}\right)^{2k\pi}\qquad or \\ \exp{\big((-|{\bf b}|+2k\pi)(2n_1J_1 + 2n_2J_2+2n_3J_3)\big)} = \left(\exp{\big(2n_1J_1 + 2n_2J_2+2n_3J_3\big)}\right)^{(-|{\bf b}|+2k\pi)} \end{array}\right. $$
Now the tricky part is to see weather $$ \exp{\big(2n_1J_1 + 2n_2J_2+2n_3J_3\big)} = E_d. $$ If you substitute vector of length 2 in the $\exp$ formula for $\mathfrak{su}(2)$ it would yield $E_2$. Now here the only difference is that $J_i$, $i = 1,2,3$ are in higher dimension, but still chosen so that they obey the same bracket relations as $\{j_i\}_{i = 1,2,3}$. So now it is all about how actually the formula (*) is derived. You can check the whole page 9 from here.
In conclusion I would like to say that many books and papers avoid direct calculations but for me this is the way to clear understanding.