Surjective function reasoning.

Question

If i have a function, $$f : X \to Y$$ Does it make sense to say the following:

For $x \in X $ and $y \in Y, $

$f$ is not surjective if $f(x) \neq y$ for some $ y \in Y$.

And $f$ is not injective if $ f(x) =f(x') $, where $x \neq x'$.

First time learning about functions in such a formal way and I don't have very detailed notes to learn off. p.s, any text recommendations which may aid my understanding about this topic and the basis of analysis in general?

Answer 1

What's really missing from your statements are quantifiers, without which it's hard to define these things. It's correct to say that $f$ is not surjective if there exists some $y$ for which there does not exist any $x$ with $f(x)=y$.

Similarly, $f$ is not injective if there exist some $x$ and $x'$ with $x\neq x'$ and $f(x)=f(x')$.

Answer 2

You're definitely grasping the concepts. You are using the contrapositive approach for your logic of injection. For instance $x\neq y\Rightarrow f(x)\neq f(y)$ so the contrapositive proof is simpler since proving equality is much more straightforward: $f(x)=f(y)\Rightarrow x=y$. So as you mention, a function is not injective if the antecedent in the contrapositive approach is true and the consequent is false.

For surjectivity I would tweak your definition to say that some $f$ is not surjective if for some $b$ in the codomain, there is no $x$ in the domain such that $f(x)=b$. Or more formally

For $f:A\rightarrow B$

$\exists b\in B,f(x)\neq b,\forall x\in A$

Also, for a good source... here is a nice overview of the theoretical side of functions: http://www.people.vcu.edu/~rhammack/BookOfProof/Functions.pdf

The exercises in the chapter are also useful.

Answer 3

To exemplify the problem in your statement "$f$ is not surjective if $(\exists x\in X) f(x)\ne y$", suppose $$f:\mathbb R\to\left\{0,1\right\}, x\mapsto 1 \text{ if }x=0,\, 0\text{ otherwise}$$ ($f$ is the indicator function of $0$). Now if I say $y=1$, then there exists $x\in\mathbb R$ such that $f(x)\ne y$ : take $x=1$. And if I say $y=0$, there is also $x\in\mathbb R$ such that $f(x)\ne y$ : take $x=0$. But $f$ is surjective.

So you see your sentence doesn't mean anything if you don't say for which $y$ you want it to be true, one, several or all (and BTW, the quantificator on $x$ must be $\forall$, not $\exists$ ! The negation of "there exists an $x$ such that $P(x)$" is "for no $x$, $P(x)$", or "for all $x$, $\neg P(x)$).