# Munkres Topology Ch. 1 section 2 Ex #1

• benorin

#### benorin

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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.

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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)).##

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?

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.

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