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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.

Rodrigo de Azevedo
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Andrew Tawfeek
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  • As a follow up question: how does a surjective function have multiple right-inverses, but I feel as if if I understand the answer to my original question that the follow up will become self-evident. – Andrew Tawfeek Mar 18 '17 at 03:08
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    A left inverse can take any value on the elements of $B$ not in the image of $f$. – Qiaochu Yuan Mar 18 '17 at 03:08
  • For right-inverses, let $A$ and $B$ be sets and $p : A \times B \to A$ be projection. Then any function $A \to A \times B$ such that any $a_0$ maps to $(a_0 , b)$ with b arbitrary is a right-inverse of $p$, since composing them takes you from $a_0$ to $(a_0,something)$ then back to $a_0$ via the projection. Notice that this needn't be a right-inverse because the choice of something in $B$ is arbitrary; we can find a bunch of functions that satisfy this. – JohnnyMo1 Mar 18 '17 at 03:37
  • Is f specifically not surjective. A bijection can 't have a multiple left inverses I think but as quaochu yuan points out if A={0,1} and B={0,1,2} and f(0)=0,f (1)=1. The both g(0)=0,g (1)=1 ,g (2)=0 and h (0)=0,h (1)=1,h (2)=1 are left inverses. – fleablood Mar 18 '17 at 06:37

2 Answers2

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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.

Sebastian Mueller
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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 $1$.

Rodrigo de Azevedo
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