The affine space of covariant derivative Let $\nabla$ be a connection on a smooth manifold $M$.
(i) Show that the set $\text{Conn}(M)$ of connections on $M$ is equal to $$\text{Conn}(M) = \left\{\nabla + A \mid A \in T_2^1(M)\right\}.$$
(ii) Let $\nabla$ be torsion-free. Show that the set of torsion-free connections $\text{Conn}(M, [\cdot, \cdot])$ on $M$ is equal to $$\text{Conn}(M,[.,.]) = \left\{\nabla + A \mid A \in T_2^1(M),~ A(X,Y) = A(Y, X)\text{ for all } X, Y ∈ X(M)\right\}.$$
(iii) Let $\nabla$ be metric with respect to a semi-Riemannian metric $g$ on $M$. Find a similar description of the set $\text{Conn}(M, g)$ of connections on $M$ which are metric with respect to $g$.
for first part, I am confused how could I get a $(1,2)$-tensor out of the connection, since connection is a map : $X\times X \to X$, $X$ is vector field. for second part and third part, I dont have any clue, could anyone show me how to prove it?
 A: I think you confusion lies in the following.
An affine connection is a map $\nabla:\mathfrak{X}(M)\times \mathfrak{X}(M)\to\mathfrak{X}(M)$ fulfilling some properties, or let's write it $\mathcal{T}_0^1(M)\times\mathcal{T}_0^1(M)\to\mathcal{T}_0^1(M)$. So it takes two $(0,1)$-tensor fields and gives back a $(0,1)$-tensor field. If you now leave open the first argument, say for fixed $Y\in\mathfrak{X}(M)$ you consider the map $$\nabla_{(\cdot)}Y:\mathcal{T}_0^1(M)\to\mathcal{T}_0^1(M),~X\mapsto \nabla_XY,$$
you find that this map is $\mathcal{C}^\infty(M)$-linear (which is one of the properties mentioned above). Hence, the resulting vector field $\nabla_XY\big|_p$ evaluated at some $p\in M$ depends only on $X\big|_p$ and on no other value of $X$ and you can consider the map
$$
\nabla_{(\cdot)}Y\big|_p:T_pM\to T_pM,~v\mapsto \nabla_{X_v}Y\big|_p
$$
with any $X_v\in\mathfrak{X}(M)$ s.t. $X_v\big|_p=v$. In other words the previous map is an element of ${T_p}_1^1(M)$ and collecting all these maps again into one field you obtain $\nabla_{(\cdot)}Y\in\mathcal{T}_1^1(M)$. Taken together you can equivalently define a connection as a map $\tilde{\nabla}:\mathfrak{X}(M)\to\mathcal{T}_1^1(M)$. Indeed some authors make it the one way, some the other, some need both versions and call one of them a covariant derivative, and some do not even make a notational difference.
Now you may notice that the lack of $\mathcal{C}^\infty(M)$-linearity in the second argument prevents you from regarding $\nabla$ as an element of $\mathcal{T}_2^1(M)$. But you can show that for two connections $\nabla,\tilde{\nabla}$ the difference $\nabla-\tilde{\nabla}:\mathfrak{X}(M)\times \mathfrak{X}(M)\to\mathfrak{X}(M)$ is $\mathcal{C}^\infty(M)$-linear in both arguments, making it indeed possible to regard it as an element $\nabla-\tilde{\nabla}\in\mathcal{T}_2^1(M)$.
Does this help you with your confusion? For your concrete problems with (ii) and (iii) you have to evaluate what the properties of being torsion-free and being metric means for the difference tensor of two connections with the respective properties.
