When is the image of a $T$-algebra under $T$ again a $T$-algebra? Let $T:\mathcal{C}\to\mathcal{C}$ be a monad with unit $\eta:1_\mathcal{C}\Rightarrow T$ and multiplication $\mu:T\circ T\Rightarrow T$, and let $(A,\alpha)$ be a $T$-algebra.

When is $\big(T(A),T(\alpha)\big)$ again a $T$-algebra?

This is trivially true when the unit is  epic and the multiplication is monic, or when $\alpha$  is monic, but this seems pretty rare.
The motivation here is that $\alpha$ would then be a $T$-algebra-homomorphism from $\big(T(A),T(\alpha)\big)$ to $(A,\alpha)$, in addition to being a structure map. This seems 'morally correct', but I can't see that it's always true off the top of my head.
 A: An object $A$ has the structure of a $T$-algebra $a\colon TA\to A$ such that $Ta\colon TTA\to TA$ is also a $T$-algebra if and only if the unit $\eta_A\colon A\to TA$ is an isomorphism, in which case $a\colon TA\to A$ is necessarily $\eta_A^{-1}$, and the multiplication $\mu_A\colon TTA\to TA$ is necessarily the isomorphism $T\eta_A^{-1}$.
For the proof, note first that a $T$-algebra is in particular a morphism $a:TA\to A$ of which $\eta_A\colon A\to TA$ is a seciton, i.e. such that $a\circ\eta_A=\mathrm{id}_A$. Let $e_{TA}=\eta_A\circ a\colon TA\to A\to TA$ be the associated split idempotent (it is an easy exercise that $e_{TA}\circ e_{TA}=e_{TA}$).
We can account at this point for the special cases of $a$ being a monomorphism or $\eta_A$ being an epimorphism. Since $a\circ e_{TA}=a\circ\eta_A\circ a=\mathrm{id}_A\circ a=a=a\circ\mathrm{id}_{TA}$ and $e_{TA}\circ\eta_A=\eta_A\circ a\circ\eta_A=\eta_A\circ\mathrm{id}_A=\mathrm{id}_{TA}\circ\eta_A$, either $a$ being an monomorphism or $\eta_A$ being an epimorphism implies $e_{TA}=\mathrm{id}_{TA}$, and hence that $a=\eta_A^{-1}$ are isomorphisms.
Back to the general case, naturality of $\eta\colon\mathrm{id}\Rightarrow T$ implies $e_{TA}=\eta_A\circ a=Ta\circ\eta_{TA}\colon TA\to TTA\to TA$. Thus for $Ta\colon TTA\to TA$ to be a $T$-algebra, and in particular for $Ta\circ\eta_{TA}=\mathrm{id}_{TA}$ to hold, it is necessary that $e_{TA}=\mathrm{id}_{TA}$, so again $a=\eta_A^{-1}$.
Conversely, $a=\eta_A^{-1}\colon TA\to A$ is a $T$-algebra if and only if $a\circ Ta=a\circ\mu_A$, which post-composition with $\eta_A$ renders  equivalent to $Ta=\mu_A$, which follows from the fact that $\mu_A\circ T\eta_A=\mathrm{id}_{TA}$ implies $\mu_A=(T\eta_A)^{-1}=T\eta_A^{-1}=Ta$.
A: The answer is that it rarely happens.
Here is a simple non-example.
Consider the so-called maybe monad on $\textbf{Set}$.
(It is the monad induced by the forgetful functor from the category of pointed sets to the category of sets.)
We might write $T X = X \amalg \{ * \}$ but this gets confusing when we iterate so (exploiting the axiom of regularity...) let me instead write $T X = X \amalg \{ X \}$.
Then $T^2 X = X \amalg \{ X \} \amalg \{ T X \}$.
An algebra structure on $T X$ is a map $\beta : X \amalg \{ X \} \amalg \{ T X \} \to X \amalg \{ X \}$ such that $\beta (x) = x$ for $x \in X \amalg \{ X \}$.
But if $\alpha : X \amalg \{ X \} \to X$ is any map whatsoever then $T \alpha (x) \ne x$ when $x = X$.
So $T \alpha$ cannot be an algebra structure on $T X$.
