Linear Transforms, Inverses, & Dimension (Question from a low level math guy): Why does it make sense that when you have a linear transformation $\textsf{T}: \mathbb{R}^n\to \mathbb{R}^{n+1}$ that you cannot have a function $\textsf{F}: \mathbb{R}^{n+1}\to \mathbb{R}^n$
that maps back you back to the vectors you started with? (like an inverse)
It makes sense as to why when you go down a dimension you can longer reach all possible points in the higher dimension using only a function with one output per input. But it doesn’t make sense to me that when you go up a dimension, you no longer can be able to map back completely to the lower dimension.
Thanks.
- SDH
 A: You seem to be looking for a more intuitive explanation so that's what I'll try to provide. To have an inverse, the given function must be injective and surjective. You made the statement:

... using only a function with one output per input.

We need to correct your use of terminology here. By definition, a function is something which assigns exactly one output per input. What you probably meant to say was something to emphasise the notion of "injective". This means if you have two different inputs, then the outputs will be different.
Using the rank-nullity theorem one can show that if $T:V \to W$ is a linear transformation which is injective, then $\dim V \leq \dim W$. This should be reasonable enough to believe intuitively. A very vague way of seeing why is:  if $T$ is injective, then it maps different inputs to different outputs, so the codomain $W$ has to be atleast "as big" as the domain $V$.
Using rank-nullity theorem, one can show that if $T: V \to W$ is linear and surjective, then $\dim V \geq \dim W$. Loosely speaking, surjective means that every element of $W$ is the output of some element of $V$ (more accurately, it means $\text{range}(T) = W$). 
You said that

...  you no longer can be able to map back completely to the lower dimension.

Recall that in general, $\text{range}(T) \subseteq W$; and if it is an equality, we say $T$ is surjective. Well, if you start from a low dimension and go to a higher one in an injective manner, you can only map back the elements in the range of $T$. Because if something is not in the range of $T$, what do you map that back to? So, the key point here is that you can only "map back" things which are in the range of $T$.
A: There will not be a function mapping all of $\Bbb R^{n+1}$ back to $\Bbb R^n,$ because a linear transformation $T$ takes $\Bbb R^n$ to a subspace of $\Bbb R^{n+1}$ with dimension at most $n,$ not to all of $\mathbb R^{n+1}$.
A: Of course it is possible to map back to $\mathbf{R}^n$, BUT, it cannot be a function on ALL of $\mathbf{R}^{n+1}$.
As I am fond of analogies, allow to to talk in a totally different context. Imagine an organization opening a bank account for paying  monthly salary to all its employees. AT year end everybody was credited a second amount, a bonus based on performance. This is function with domain employees of the organization codomain being customers of the bank.
Now HR tells the account department due to some bug in the software bonus was calculated wrongly.
To undo the mistake  you have to map back the same amount from the employees account.
But originally we had defined the function as having ALL customers of that bank as codomain.
The employer cannot take money frm them all, only those who work for them.
Now coming back to the original question inverse is NOT defined on all of the codomain, only on those vectors which are in the range of $T$, so it is not possible to define $F$ as a mathematical function on $\mathbf{R}^{n+1}$ as the domain. Though possible to define  on a restricted subset.
A: As long as $T : \mathbb R^{n} \to \mathbb R^{n+1}$ is injective, you certainly can define a linear map $F : \mathbb R^{n+1} \to \mathbb R^n$ which is a left inverse of $T$, so that $F(T(x)) = x$ for all $x \in \mathbb R^n$.
To construct such an $F$, start by defining $F(y) = x$ for every $y \in \operatorname{image}(T)$, where $x$ is the unique element of $\mathbb R^{n}$ such that $T(x) = y$.
Next, let $B = \{b_1,b_2,\ldots,b_n\}$ be any basis for $\operatorname{image}(T)$. Extend $B$ to a basis of $\mathbb R^{n+1}$ by appending any vector $v \in \mathbb R^{n+1}$ that is linearly independent of $B$. Define $F(v)$ to be any desired element of $\mathbb R^n$. Then observe that an arbitrary $w \in \mathbb R^{n+1}$ can be expressed uniquely as $w = y + \alpha v$, where $y \in \operatorname{image}(T)$ and $\alpha \in \mathbb R$. Define $F(w) = F(y) + \alpha F(v)$. It's easy to check that $F$ is linear and satisfies $F(T(x)) = x$ for all $x \in \mathbb R^{n}$.
