Are these functions isomorphisms of the first binary structure with the second? - Fraleigh p.34 3.13, 3.15 
(1.) How can we suspect these two aren't isomorphisms, before trying to find counterexamples?
 (2.) I  don't understand their arguments. 
For 13, how is there no $x$ such that $\begin{align} \phi(f)(x) & = x + 1 \\ \int^{x}_{0} f(t) dt &= \end{align}$?
For 15, how is there no $x$ such that $\begin{align} \phi(f)(x) &= 1 \\ x\cdot f(x)& = \end{align}$ ? Don't know what $f(x)$ is?
 A: I am posting about $15$. What we are looking for $15$. Indeed, we want to examine the map $\phi$ for being an isomorphism. Any such that map should be a well-defined homomorphism and be a bijection form $F$ to $F$. 


*

*it is well-defined function:


$$f(x)=g(x)\Longrightarrow \phi [f(x)]=x\cdot f(x)=x\cdot g(x)=\phi[g(x)]$$ 


*

*It is an homomorphism between groups:


$$\phi(f)\phi(g)(x)=\phi(f)\left(x\cdot g(x)\right)=x\cdot f(x)\cdot g(x)=x\cdot(fg)(x)=\phi(fg)(x)$$


*

*Is it one-one map? If it wants to be like that, we should verify the following statement:


$$\phi(f)=\phi(g)\Longrightarrow f=g$$ If $\phi(f)=\phi(g)$ so every real $x$ we have $\phi(f)(x)=\phi(g)(x)$, so we have $$x\cdot f(x)=x\cdot g(x)$$ Can we cancel that irritating $x$ form both sides? No! Unless we assume $x\neq 0$. This is the point that counterexample appears. In fact the values of $\phi$ when acting on every two non-equal functions $f$ and $g$ in $F$ while $x=0$ are the same. Therefore $\phi$ cannot satisfy the injectivity and so it is not an isomorphism. If we want to work the sample the book gave us, we can say:
Let $\phi$ is an isomorphism. So it is onto from $F$ to $F$. So for example, if we set $\color{darkcyan}{f=1}\in \langle F,\cdot\rangle$ in the right hand side, then we should have $g\in\langle F,\cdot\rangle $ on left hand side such that $$\phi(g)(x)=\color{darkcyan}{f=1}, ~~x\in\mathbb  R$$ Now what would be happen if we have $$\phi(g)(x)=x\cdot g(x)=1$$ while $x=0$?? Indeed, we would get: $$0=\phi(g)(0)=0\cdot g(0)=1$$
This is a nice contradiction.
A: HINTS:


*

*For 13, compute the value of $\phi(f)(0)$ for any $f$ whatsoever. Does $x+1$ have this value at $0$ as well?

*For 15, compute the value of $\phi(f)(0)$ for any $f$ whatsoever. Does constant $1$ function have this value at $0$ as well?
A: B.S. answered 15 nicely and exquisitely. I try to do the sae for 13. 
• Is it a well-defined function? Yes. I can see this.
• Is it an homomorphism between groups:
$\begin{align}( \, \phi(f) + \phi(g) \, )(x) & = \int_{[0, x]}f(t) + g(t) \, dt \\ & = \int_{[0, x]}f(t) \, dt + \int_{[0, x]} g(t) \, dt = \phi(f)(x) + \phi(g)(x) \end{align}$. Yes.
• Is it one-to-one map? $\phi(f)=\phi(g)\Rightarrow \color{purple}{\frac{d}{dx}} \int_{[0, x]}f(t) \, dt = \color{purple}{\frac{d}{dx}} \int_{[0, x]}g(t) \, dt \implies f(x) = g(x) $. Yes.
• Is it onto map? For all $r(x) \in <F, +>,$ is there $f(x)$ such that $\phi(f)(x) = r(x)$? 
Start from what you need to prove to find $f(x)$. $\phi(f)(x) = r(x) \implies \int_{[0, x]}f(t) \, dt = r(x) \implies \color{purple}{\frac{d}{dx}}  \int_{[0, x]}f(t) \, dt = r(x) \implies f(x) = r'(x) $. 
Yes. 
Where's the problem? 
