How could an injective function have multiple left-inverses? Maybe I'm a bit tired, but I can't seem to imagine two different left-inverses for an injective function. This was brought up in Aluffi's book.
To quote it nearly verbatim, if $f$ is a function going $A \to B$ and $A$ has at least two elements, then it will necessarily have more than one left-inverse.
Both concrete examples and more abstract/general explanations are very welcome.
 A: The problem is that your quote is nearly verbatim. You forgot an important point (I just took the effort to look into his book to see the precise statement).
He said that

If a function is injective but not surjective, then it will not have a right inverse, and it will necessarily have more than one left inverse

The important point being that it is NOT surjective. This means that there is a $b\in B$ such that there is no $a\in A$ with $f(a) = b$. When defining a left inverse $g: B \longrightarrow A$ you can now obviously assign any value you wish to that $b$ and $g$ will still be a left inverse. Thus, as long as $A$ has more than one element, there is a multitude of such functions. 
A: Let injective function $\mathrm v : \mathbb R \to \mathbb R^3$ be defined by
$$\mathrm v (t) = t \begin{bmatrix} 1\\ 1\\ 1\end{bmatrix}$$
and let $f : \mathbb R^3 \to \mathbb R$ be defined by
$$f (\mathrm x) = \begin{bmatrix} 1\\ 0\\ 0\end{bmatrix}^{\top} \mathrm x$$
Since,
$$(f \circ \mathrm v) (t) = t \, \begin{bmatrix} 1\\ 0\\ 0\end{bmatrix}^{\top} \begin{bmatrix} 1\\ 1\\ 1\end{bmatrix} = t$$
for all $t \in \mathbb R$, then $f$ is a left inverse of $\mathrm v$. However, function $g : \mathbb R^3 \to \mathbb R$, which is defined by
$$g (\mathrm x) = \frac{1}{3} \begin{bmatrix} 1\\ 1\\ 1\end{bmatrix}^{\top} \mathrm x$$
is another left inverse of $\mathrm v$, for $(g \circ \mathrm v) (t) = t$ for all $t \in \mathbb R$. There are infinitely many left inverses of $\mathrm v$ because there are infinitely many vectors whose inner product with $\begin{bmatrix} 1 & 1 & 1\end{bmatrix}^{\top}$ is $1$.
