Munkres Topology Ch. 1 section 2 Ex #1

[benorin]

Homework Statement

If $f: A\rightarrow B$ and $A_0\subset A$ and $B_0\subset B$. (a) show that $A_0\subset f^{-1}(f(A_0))$ and that equality holds if $f$ is injective. (b) show that $f(f^{-1}(B_0))\subset B_0$ and that equality holds if $f$ is surjective.

Relevant Equations

The definitions of injective and surjective. Also definitions of $f$ and $f^{-1}$ restricted to a subset of the domain or range. I think I understand this part on my own but will type these up below if you want me to, I’m on my phone and TeX is a pain.

Mostly I need to clear up a few basic things about functions and their inverses, the problem seems easy enough. Ok, so for (a) I have

$$f^{-1}(f(A_0))= \left\{ f^{-1}(f(a)) | a\in A_0\right\}$$

but here I’m not certain if $f^{-1}$ is allowed to be multi-valued or not, the text says that if $f$ is bijective then $f^{-1}$ exists and is also bijective. But it didn’t specifically say “if, and only if” just if. And clearly the problem stipulates the existence of the inverse of $f$ even if it is not even injective so I’m unclear as to the nature of the inverse function here? I hope that makes sense.

 

[Infrared]

If $f$ is a function from $A$ to $B$ and if $C\subset B$, then $f^{-1}(C)$ means, just by definition, the set $\{a\in A: f(a)\in C\}.$ It is not the image of $C$ under a function called $f^{-1}$, and we are not assuming that $f$ is invertible.

With this in mind,

$$f^{-1}(f(A_0))= \left\{ f^{-1}(f(a)) | a\in A_0\right\}$$

doesn't make sense. In your RHS, you are applying $f^{-1}$ to an element of $B$ rather than a subset of $B$. Even if you were to write $\{f^{-1}(\{f(a)\}):a\in A_0\}$ on your RHS (and it is indeed a common 'abuse of notation' to write $f^{-1}(x)$ in place of $f^{-1}(\{x\})$), it would still be off. For example, suppose $f:\{0,1\}\to \{0,1\}$ is just the constant function $f(0)=f(1)=0.$ Then your RHS is $\{f^{-1}(0)\}=\{\{0,1\}\},$ which is not a subset of the domain, whereas your LHS is $f^{-1}(\{0\})=\{a\in \{0,1\}: f(a)\in\{0\}\}=\{a\in \{0,1\}: f(a)=0\}=\{0,1\}.$

A correct expression along these lines would be $f^{-1}(f(A_0))=\bigcup_{a\in A_0} f^{-1}(\{f(a)\}),$ but this doesn't really get you closer to a solution. Instead, try to show that $A_0$ is always a subset of $f^{-1}(f(A_0))$ by taking an element of $A_0$ and showing that it must be in $f^{-1}(f(A_0)).$

 

[benorin]

If $f$ is a function from $A$ to $B$ and if $C\subset B$, then $f^{-1}(C)$ means, just by definition, the set $\{a\in A: f(a)\in C\}.$ It is not the image of $C$ under a function called $f^{-1}$, and we are not assuming that $f$ is invertible.

This^ helps, thanks. Now that I'm on my Mac (and not my phone) I will write the step I skipped earlier, namely

$$\boxed{f^{-1}(f(A_0))= \left\{ f^{-1}(b) | b\in f(A_0)\right\} } =\left\{ f^{-1}(f(a)) | a\in A_0\right\}$$
I now know that the second equality doesn't make sense given what you said. I did think that $\{ b | b\in f(A_0)\} = \{ f(a) | a\in A_0\}$ was an ok simplification. I guess not.

With this in mind,doesn't make sense. In your RHS, you are applying $f^{-1}$ to an element of $B$ rather than a subset of $B$. Even if you were to write $\{f^{-1}(\{f(a)\}):a\in A_0\}$ on your RHS (and it is indeed a common 'abuse of notation' to write $f^{-1}(x)$ in place of $f^{-1}(\{x\})$), it would still be off. For example, suppose $f:\{0,1\}\to \{0,1\}$ is just the constant function $f(0)=f(1)=0.$ Then your RHS is $\{f^{-1}(0)\}=\{\{0,1\}\},$ which is not a subset of the domain, whereas your LHS is $f^{-1}(\{0\})=\{a\in \{0,1\}: f(a)\in\{0\}\}=\{a\in \{0,1\}: f(a)=0\}=\{0,1\}.$

A correct expression along these lines would be $f^{-1}(f(A_0))=\bigcup_{a\in A_0} f^{-1}(\{f(a)\}),$ but this doesn't really get you closer to a solution. Instead, try to show that $A_0$ is always a subset of $f^{-1}(f(A_0))$ by taking an element of $A_0$ and showing that it must be in $f^{-1}(f(A_0)).$

From the usage in the text, I've inferred that $f^{-1}$ may be applied to both an element of $B$ and a subset of $B$, where context is key, as I have seen both explicitly; is the former only valid if $f$ is invertible?

 

[Infrared]

I did think that $\{ b | b\in f(A_0)\} = \{ f(a) | a\in A_0\}$ was an ok simplification. I guess not.

This is basically correct, but not what you did in your OP. The LHS is a little clunky though- why would you write $\{x: x\in E\}$ instead of just $E.$?

In general though, when writing sets in this way, it's better to write something like $\{x\in C: \Phi(x)\}$ instead of $\{x:\Phi(x)\}$ where $\Phi(x)$ is some predicate. If you don't include a domain, then you can get issues like Russel's paradox (thinking about the "set" $\{x:x\notin x\}$ leads to contradictions) as well as just annoying ambiguities, e.g. in $\{x: x^3=1\}$, is $x$ supposed to be a real number, or a complex number, or maybe a matrix...? Here it's clear what you mean, but just be careful.

From the usage in the text, I've inferred that $f^{-1}$ may be applied to both an element of $B$ and a subset of $B$, where context is key, as I have seen both explicitly; is the former only valid if $f$ is invertible?

If $f$ is invertible, then $f^{-1}(b)$ makes sense- it is just the function $f^{-1}$ applied to the element $b.$ If $f$ is not necessarily invertible, then to be perfectly strict, you should only write $f^{-1}(C)$ where $C$ is a subset of $B$, but sometimes $f^{-1}(b)$ is written as shorthand for $f^{-1}(\{b\})$, and this latter expression is well-defined (it's the set of all elements of $A$ that map to $b$).
For example, if $f:\mathbb{R}\to\mathbb{R}$ is the function $f(x)=x^2$, then $f^{-1}(1)=f^{-1}(\{1\})=\{1,-1\}$ according to this notation. $f^{-1}$ is not a function on its own. It might be a good idea to avoid this shorthand if you're still getting used to this stuff.