Context: Seth Warner's "Modern Algebra" (1965), exercise $13.13$. Ongoing self-study.
Let $(S, \odot)$ and $(T, \otimes)$ be closed algebraic structures with one operation. Let $(S, \odot)$ be such that if $f: T \to S$ and $g: T \to S$ are homomorphisms, then $f \odot g$ is also a homomorphism. Then $(S, \odot)$ is necessarily an entropic structure.
EDIT: After a comment by Eric Wofsey, I look more closely and it should say:
Let $(S, \odot)$ be a closed algebraic structures with one operation. Let $(S, \odot)$ be such that if $(T, \otimes)$ is any arbitrary closed algebraic structure with one operation, and $f: T \to S$ and $g: T \to S$ are homomorphisms, then $f \odot g$ is also a homomorphism.
In the above:
$f \odot g$ denotes the pointwise operation induced by $\odot$, that is: $(f \odot g) (x) = f(x) \odot g(x)$.
By "entropic", I mean that $(w \odot x) \odot (y \odot z) = (w \odot y) \odot (x \odot z)$ for all $w, x, y, z \in S$.
a "closed algebraic structure" is simply a "magma".
I have worked through the straightforward algebra as follows:
$(f (a) \odot f (b)) \odot (g (a) \odot g (b))$
$=(f (a \otimes b) )\odot (g (a \otimes b) )$ (as $f$ and $g$ are both homomorphisms)
$= (f \odot g) (a \otimes b)$ (by definition of pointwise operation)
$= (f \odot g) (a) \odot (f \odot g) (b)$ (as $f \odot g$ is a homomorphism)
$= (f (a) \odot g (a) \odot (f (b) \odot g (b))$ (by definition of pointwise operation)
Thus we have: $$(f (a) \odot f (b)) \odot (g (a) \odot g (b)) = (f (a) \odot g (a) \odot (f (b) \odot g (b))$$
and thus $(S, \odot)$ exhibits the symptoms of being entropic.
But this only demonstrates that elements in the image set of $f$ and $g$ have the property that make $\odot$ entropic. From the above analysis, it is entirely possible that there exist $w, x, y, z \in S$ which are not the images of homomorphisms, which do not have the property that $(w \odot x) \odot (y \odot z) = (w \odot y) \odot (x \odot z)$.
How do I rescue this proof?
(Note of course that based on the above, the converse is trivial to prove: that if $(S, \odot)$ is entropic, then for homomorphisms $f$ and $g$ from $T$ to $S$ it follows that $f \odot g$ is also necessarily a homomorphism.)
Please feel free to add whatever tags are necessary for this question -- I haven't managed to find any more than the one.
This is not true as written. For instance, if $T$ is empty, then no matter what $S$ is, there is only one map $T\to S$, and so the hypothesis is always trivially true, regardless of whether $S$ is entropic. Or, if $T$ is a singleton and $S$ has no idempotent elements, there are no homomorphisms $T\to S$, so the hypothesis holds vacuously, but $S$ does not have to be entropic (for instance, it could be the free magma on one element).
I would guess that the statement was intended to require that the hypothesis hold for all $T$, rather than just one fixed $T$. In that case, you can prove it by taking $T$ to be the free magma on two generators $a$ and $b$. Then for any $w,x,y,z\in S$, by the universal property of $T$ there exists a homomorphism $f:T\to S$ sending $a$ and $b$ to $w$ and $x$ and a homomorphism $g:T\to S$ sending $a$ and $b$ to $y$ and $z$. Combined with your argument this shows that $S$ must be entropic.