Exercise about algebras for a monad 
The forgetful functor $U:\bf{ Top}\to \bf{Set}$ is a left adjoint. If $I$ denotes its right adjoint,  describe the category of the algebras for the associated monad $T=IU$.

First off, the right adjoint is the functor that sends a topological space $(X,\tau)$ to $(X,\iota)$, where $\iota$ denotes the trivial topology. It follows that $T(X,\tau)=(X,\iota)$, $\eta_X:(X,\tau)\to(X,\iota)$ is (the continuous function with underlying map) the identity on $X$, and $\mu_X:(X,\iota)\to (X,\iota)$ again the identity. I don't understand what are the algebras $((X,\tau),h)$ for this monad, since by the unit law I have that the underlying map of $h$ is the identity on $X$; however this is not continuous as a function $(X,\iota)\to (X,\tau)$. I would say that the continuous functions $(X,\iota)\to (X,\tau)$ are all the (set-theoretic) maps whose image is the closure of a point in $(X,\tau)$. I don't understand if it means that the $T$-algebras does not "add structure" in this case, but "add properties", in the sense that they are not all topological spaces but only the spaces that are the closure of a point, or if I'm just wrong. Thanks for any clarify
 A: Suppose $(A,\alpha)$ is an algebra for $T$. Then $A$ is a topological space $(X,\tau)$ and $\alpha\colon T(A)\to A$ is a continuous map from $(X,\iota)$ to $(X,\tau)$, where $\iota$ is the trivial topology on $X$, as in your question.
We also know that $\alpha\circ \eta_A = \text{id}_A$, by the unit diagram. Since $\eta_A$ is the map $T(A) = (X,\iota) \to A = (X,\tau)$ which is the identity on points, and $\text{id}_A$ is also the identity on points, $\alpha$ must be the identity on points as well. Since $\alpha\colon (X,\iota)\to (X,\tau)$ is continuous, it follows that every open set in $\tau$ is open in $\iota$, and hence $\tau$ is the trivial topology.
What we've shown is that if $(A,\alpha)$ is an algebra for $T$, then $A$ is a topological space with the trivial topology, so $T(A) = A$, and $\alpha$ is the identity map. Conversely, it's not hard to check that whenever $A$ is a topological space with the trivial topology, $(A,\text{id}_A)$ is an algebra for $T$. So the algebras for $T$ are exactly the topological spaces with the trivial topology.
This is what's meant by only "adding properties". The base category is the category of topological spaces. A $T$-algebra is not a topological space with extra structure, but rather a certain kind of topological space. Put another way, the category of $T$-algebras is a full subcategory of the category of topological spaces.
