Showing that a map is not an isotopy In this wikipedia example we read:

The map from the interval $[-1,1]$ into the real numbers defined by $f(x) = -x$ is not isotopic to the identity $g(x)=x$.  Any homotopy from $f$ to the identity would have to exchange the endpoints, which would mean that they would have to 'pass through' each other. Moreover, $f$ has changed the orientation of the interval and $g$ has not, which is impossible under an isotopy. However, the maps are homotopic; one homotopy from $f$ to the identity is $H:[-1,1]\times[0,1];→[-;1,1]$ given by $H(x,t)=(2t-1)x$.

I want to understand why the defined homotopy $\forall t\in [0,1]$, $H_t(x)=(2t-1)x$ is not an isotopy. My understanding of isotopy is that we require that $\forall t\in [0,1]$ the map $H_t$ to be continuous and injective. I can see that $H_{1/2}$ is the zero map so it is not an injective map, hence $H$ can't be an isotopy. But i can't understand the meaning of the sentences :
1) "would have to exchange the endpoints"
2)"they would have to 'pass through' each other"
3) "$f$ has changed the orientation of the interval and $g$ has not"
Thank you for your help!!
 A: An isotopy is a homotopy from $f$ to $g$ for which each of paths intermediate to $f$ and $g$ is an embedding.  But before we go further, let's back up a bit.
If $f,g:X\to Y$ are two continuous functions, then a continuous function $H:[0,1]\times X \to Y$ is a homotopy from $f$ to $g$ if 
$$\forall x \in X \, \big(H(0,x)=f(x) \textrm{ and } H(1,x)=g(x)\big).$$
If such a homotopy exists, then we say that $f$ can be continuously deformed into $g$. 
So far, so good.  I think that you already knew all that. 
Now, if $f$ and $g$ are embeddings (which in this case just means continuous injections) an isotopy is a homotopy $H$ such that for each $t\in [0,1]$ the function $x\mapsto H(t,x)$ is also an embedding. 
In our case, $X$ is the interval $[-1,1]$ and $\mathbb{R}$ is $Y$ (although we could just as well use $[-1,1]$).  the embedding $f$ is the function $x\mapsto -x$, and $g$ is the identity function on $[-1,1]$. 
Now, think about a homotopy from $f$ to $g$.  As $t$ goes from $0$ to $1$, $H(t,-1)$ must go from $1$ to $-1$.  Likewise, $H(t,1)$ must go from $-1$ to $1$.  Since these are both continuous functions of $t$, they must, by the intermediate value theorem (or Rolle's theorem, if you prefer), cross: there must be some $t_{0}\in [0,1]$ such that $H(t_{0},-1)=H(t_{0},1)$. But it must therefore be the case that $x\mapsto H(t_{0},x)$ is not injective, and so not an embedding.  $H$ is therefore not an isotopy. 
It is the existence of this $t_{0}$ that the text is referring to when it says that we would have to "exchange the endpoints." Hopefully, you can also visualize the "crossing through" that is described, and how that crossing makes it impossible for the intermediate functions to be embeddings.   
