Puzzling question in Warner's "Modern Algebra" -- isomorphism from $(R, \cdot)$ to $(R, \cdot)$ Seth Warner's "Modern Algebra" exercise $6.6$ is as follows:
"Let $\vee$ and $\wedge$ be the compositions on $\mathbb R$ defined by $$x \vee y = \max \{x, y\}$$ $$x \wedge y = \min \{x, y\}$$
(a) Exhibit an isomorphism from $(\mathbb R, \vee)$ onto $(\mathbb R, \wedge)$ and one from $(\mathbb R, \cdot)$ to $(\mathbb R, \cdot)$. (b) Prove that $(\mathbb R, \vee, \cdot)$ is not isomorphic to $(\mathbb R, \wedge, \cdot)$. ..."
(Note that in the above, $\mathbb R$ is used to denote the real numbers.)
In (a), The first isomorphism is simple enough -- the isomorphism in question is $\phi(x) := -x$.
But can the answer to the second of these really be as trivial as the identity mapping? Does Warner really want us to find an isomorphism to map a semigroup to itself?
(I appreciate that the object of the exercise is probably to demonstrate that 2 algebraic structure of 2 operations may not be isomorphic despite the fact that the corresponding structures of 1 operation are isomorphic, but because it has been established that every alg. struct. is isomorphic to itself via the identity mapping, I wonder why he should have seen fit to get the student to work through this all over again.)
EDIT: Several people are providing answers to part (b) above, which is all very good, but I was specifically asking about the second part of question (a), where we are asked to find an isomorphism from $(\mathbb R, \cdot)$ to $(\mathbb R, \cdot)$. I've got (b) covered; it's straightforward and simple. But as "find an isomorphism from $(\mathbb R, \cdot)$ to $(\mathbb R, \cdot)$" is so obvious and trivial, I am at a loss to understand why it would have been set as an exercise.
 A: You are correct that the identity mapping is always an isomorphism and is the simplest (though not the only) isomorphism $(\mathbb{R}, \cdot) \to (\mathbb{R}, \cdot)$. Other isomorphisms exist.
The continuous isomorphisms are all of the form
$$f(x) = \begin{cases}
  0 & x = 0 \\
  x^j & x > 0 \\
  -|x^j| & x < 0
\end{cases}$$
for some positive $j$. If you take $j$ negative, you can get some other isomorphisms. Other extremely non-continuous isomorphisms exist, but it appears to require the axiom of choice to prove it.
Edit: it does indeed require choice to prove there are other isomorphisms. For consider that the property “$x$ is positive” can be written as $\exists y \forall z (x \cdot y \cdot y \cdot z = z)$. Thus, an isomorphism $(\mathbb{R}, \cdot) \to (\mathbb{R}, \cdot)$ must restrict to an isomorphism $(\mathbb{R}_+, \cdot) \to (\mathbb{R}_+, \cdot)$. Now $(\mathbb{R}_+, \cdot)$ is isomorphic to $(\mathbb{R}, +)$, and it is consistent with ZF + dependent choice that the only group homomorphisms $(\mathbb{R}, +) \to (\mathbb{R}, +)$ are linear - see https://mathoverflow.net/questions/57426/are-there-any-non-linear-solutions-of-cauchys-equation-fxy-fxfy-witho/57532#57532.
For the second case, we are trying to show that the two structures are not isomorphic.
To show that they are not isomorphic, consider the language containing $*$ and $\cdot$ where $*, \cdot$ are binary operators.
Consider the statement $\exists x \exists z, [\forall y, y \cdot x = y \land y \cdot z = z)] \land (x * z = z)$.
Note that when we interpret this statement in a structure $(\mathbb{R}, *, \cdot)$, the $x$ and $z$ in question must satisfy $x = 1$ and $z = 0$. So the statement boils down to whether $1 * 0 = 0$. This is a true statement when $* = \land$ but a false statement when $* = \lor$.
Therefore, the statement is true in the model $(\mathbb{R}, \land, \cdot)$ but false in the model $(\mathbb{R}, \lor, \cdot)$. So the two models cannot be elementarily equivalent, hence cannot be isomorphic.
To phrase the above in purely algebraic terms, suppose there were an isomorphism $f : (\mathbb{R}, \land, \cdot) \to (\mathbb{R}, \lor, \cdot)$. I claim that it must be the case that $f(1) = 1$. For take $k$ such that $f(k) = 1$. Then $1 = f(k) = f(1 \cdot k) = f(1) \cdot f(k) = f(1) \cdot 1 = f(1)$.
Similarly, it must be the case that $f(0) = 0$. Take some $k$ such that $f(k) = 0$. Then we have $f(0) = f(0 \cdot k) = f(0) \cdot f(k) = f(0) \cdot 0 = 0$.
Then we have $0 = 1 \land 0$. Applying $f$ to both sides gives us $0 = f(0) = f(1 \land 0) = f(1) \lor f(0) = 1 \lor 0 = 1$, which is a contradiction.
As far as pedagogical purpose, the point is to show that you can’t “reverse quantifiers” in this context.
In other words, you can’t go from “for all function symbols $*$, there is a function preserving $*$” to “there is a function $f$ such that for all function symbols $*$, $f$ preserves $*$”. You can’t go from $\forall * \exists f$ to $\exists f \forall *$.
People in intro math classes mess this up all the time. It’s an extremely common error.
A: The second question is asking to show there is no bijection $\rho:\mathbb R\rightarrow\mathbb R$ such that $$\forall x,y\in\mathbb R, \rho(x\vee y) = \rho(x)\wedge \rho(y) \text{ and } \rho(xy) = \rho(x)\rho(y)$$
That is $\rho$ has to be compatible with both operations at the same time. We cannot pick a $\rho_1$ for $\wedge/\vee$ and then $\rho_2$ for multiplication.
Suppose there is such a $\rho$, and pick $\rho(x)=2$ (which exists as $\rho$ is surjective), then $$\rho(0)=\rho(x\cdot 0) = \rho(x) \cdot \rho(0) = 2\rho(0)$$ hence $\rho(0)=0$.
We have that $\rho(1) = \rho(1^2\vee 0) = \rho(1^2)\wedge \rho(0) = \rho(1)^2 \wedge 0 = 0$. Therefore $\rho(0)=\rho(1)=0$, so $\rho$ is not injective, a contradiction.
