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Let $f:A\to B$ be a function with $A_i\subset A, i=0,1$. I want to understand the precise logic behind the need for injectivity for $f(A_0\setminus A_1)=f(A_0)\setminus f(A_1)$ to be true.

First we prove $f(A_0)\setminus f(A_1)\subset f(A_0\setminus A_1)$ since this direction does not require injectivity:

Let $b\in f(A_0)\setminus f(A_1)$. This is equivalent to $b\in f(A_0)$ and $b\notin f(A_1)\iff f^{-1}(b)\in A_0$ and $f^{-1}(b)\notin A_1$. The latter implies, but is not equivalent to, $f^{-1}(b)\in(A_0\setminus A_1)\iff b\in f(A_0\setminus A_1).$ Thus $f(A_0)\setminus f(A_1)\subset f(A_0\setminus A_1)$.

Note here $f^{-1}(b)$ does not refer to a necessarily unique element, just an $a\in A$ such that $f(a)=b$. So therefore $f^{-1}(b)\in A_0$ and $f^{-1}(b)\notin A_1$ only implies $f^{-1}(b)\in(A_0\setminus A_1)$ since if $f^{-1}(b)\in(A_0\setminus A_1)$ were to imply $f^{-1}(b)\in A_0$ and $f^{-1}(b)\notin A_1$ we would find in case $f(a_0)=f(a_1)=b$ with $A_0=\{a_0\}, A_1=\{a_1\}$ that $$a_0=f^{-1}(b)\in(A_0\setminus A_1)\implies a_1=f^{-1}(b)\in A_0 \text{ and } a_1=f^{-1}(b)\notin A_1$$ which is false.

Now we prove $f(A_0\setminus A_1)\subset f(A_0)\setminus f(A_1)$ if $f$ is injective:

Let $b\in f(A_0\setminus A_1).$ This is equivalent to $f^{-1}(b)\in (A_0\setminus A_1).$ But since $f$ is injective, the latter $f^{-1}(b)\in (A_0\setminus A_1)$ is equivalent to $f^{-1}(b)\in A_0$ and $f^{-1}(b)\notin A_1$ since $f^{-1}(b)$ is the unique $a\in A$ such that $f(a)=b.$ So we then have $b\in f(A_0)$ and $b\notin f(A_1)$, or $b\in f(A_0)\setminus f(A_1).$ Thus $f(A_0\setminus A_1)\subset f(A_0)\setminus f(A_1)$.

So I guess my question is, is my invocation of injection for a unique $a$ really "enough" per se without that middle explanatory paragraph, or should I be do something like Robert Cardona does here to more directly use the injectivity condition? I use the exact same minimal justification of uniqueness of the inverse in my proof of $f(A_0\cap A_1)=f(A_0)\cap f(A_1)$ too so I'm wanting to get this right.

The thing that's interesting is that injectivity turns $f^{-1}(b)$ into an element, whereas generally for non injective functions $f^{-1}(b)$ denotes a set. So therefore, you can do regular set theory containment without worrying about quantifiers when $f$ is injective.

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    You need injectivity for $f(A\cap B)=f(A)\cap f(B)$ to be true (and thus, for $f(A\setminus B)=f(A)\setminus f(B)$ to be true). Notice that you can't write $f^{-1}(b)$ since this is not defined if $f$ is not bijective. – Surb May 09 '22 at 09:19
  • @Surb Yes I know...I'm basically just asking if my way of incorporating the injectivity condition in the second proof is correct. The point I was making at the end there which you touched on in your comment is that it seems if you start by assuming $b\in f(A_0\setminus A_1)$ i.e. in the image of the function and if $f$ is injective, you can treat $f^{-1}(b)$ as an element and proceed with the proof like you would "think" or "like to" via element chasing. –  May 09 '22 at 09:26

1 Answers1

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If $f$ is not injective, then the claim $f(A_0\setminus A_1)\subseteq f(A_0)\setminus f(A_1)$ may not be true.

For example, take

$$f:\{1,2\}\to \{1,2\}$$

defined as $f(1)=1, f(2)=1$. Also, take $A_0 = \{1,2\}$ and $A_1=\{1\}$.

Then: $$f(A_0\setminus A_1) = f(\{2\}) = \{1\}$$

while $$f(A_0)\setminus f(A_1) = f(\{1,2\})\setminus f(\{1\}) = \{1\}\setminus\{1\} = \emptyset\neq \{1\}=f(A_0\setminus A_1)$$

Taroccoesbrocco
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5xum
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  • I know...I'm asking whether the way I incorporated the injectivity condition in my second proof of the $f(A_0\setminus A_1)\subset f(A_0)\setminus f(A_1)$ direction is sufficient. –  May 09 '22 at 09:23
  • @upanddownintegrate It is hard to say because your proof goes off the rails before that. Talking about $f^{-1}(b)$ when $f$ is not bijective is an error, and it muddles everything else that is written. I strongly recommend you thoroughly rewrite both proofs to avoid such an error. – 5xum May 09 '22 at 09:34