Confused about Kleisli category target object (codomain) According to Wikipedia Kleisli category is defined as:
\begin{aligned}{\mathrm  {Obj}}({{\mathcal  {C}}_{T}})&={\mathrm  {Obj}}({{\mathcal  {C}}}),\\{\mathrm  {Hom}}_{{{\mathcal  {C}}_{T}}}(X,Y)&={\mathrm  {Hom}}_{{{\mathcal  {C}}}}(X,TY).\end{aligned}
Where I'm confused: $Hom_{C_T}(X,Y)$ The codomain here is $Y$ which is obviously wrong since the real codomain is actually $TY$. Isn't this like cheating claiming an incorrect codomain?
Edit:
I think I got the source of my confusion: a morphism has a source and a target whereas functions have domains and codomains. Category theory is just more abstract and when transfering concepts from Sets/functions to category theory one has to take account of that.
 A: No. The codomain of a morphism from $X$ to $Y$ in the Kleisli category $C_T$ is $Y$ by definition. It corresponds to, but is not the same thing as, a morphism from $X$ to $T(Y)$ in the original category $C$, which has codomain $T(Y)$. 
This definition is also incomplete: it hasn't told you how to compose Kleisli morphisms, which you do using the monad structure of $T$. Specifically, if $f : X \to Y$ and $g : Y \to Z$ are two morphisms in the Kleisli category, they correspond to (but are not the same things as) two morphisms $f' : X \to T(Y)$ and $g' : Y \to T(Z)$. The composite Kleisli morphism is the morphism in $C_T$ corresponding to (but not the same thing as) the composite
$$X \xrightarrow{f'} T(Y) \xrightarrow{T(g')} T^2(Z) \xrightarrow{m(Z)} T(Z)$$
in $C$, where $m : T^2 \to T$ is the monad multiplication. 
It would be a useful exercise to work through what this means very explicitly in the case that $C = \text{Set}$ and $T$ is the list monad. Then you can think of a Kleisli morphism $X \to Y$ as a function which, given an element of $X$, returns a list of elements of $Y$, and I'll leave you to work out what Kleisli composition is. 
