## 1.2 Exercise 2

"Let $f : A \rightarrow B$  and let $A_i \subset A$  and $B_i \subset B$  for $i = 0$  and $i = 1$.  Show that $f^{-1}$  preserves inclusions, unions, intersections, and differences of sets:

(a) $B_0 \subset B_1 \Rightarrow f^{-1}(B_0) \subset f^{-1}(B_1)$

(b) $f^{-1}(B_0 \cup B_1) = f^{-1}(B_0) \cup f^{-1}(B_1)$

(c) $f^{-1}(B_0 \cap B_1) = f^{-1}(B_0) \cap f^{-1}(B_1)$

(d) $f^{-1}(B_0 \setminus B_1) = f^{-1}(B_0) \setminus f^{-1}(B_1)$

Show that $f$  preserves inclusions and unions only:

(e) $A_0 \subset A_1 \Rightarrow f(A_0) \subset f(A_1)$

(f)  $f(A_0 \cup A_1) = f(A_0) \cup f(A_1)$

(g) $f(A_0 \cap A_1) \subset f(A_0) \cap f(A_1)$ ; show that equality holds if $f$  is injective.

(h) $f(A_0 \setminus A_1) \supset f(A_0) \setminus f(A_1)$ ; show that equality holds if $f$  is injective."

(Taken from Topology by James R. Munkres, Second Edition, Prentice Hall, NJ, 2000. Page 20.)

NB:  A word about notation.  $f^{-1}(B_i)$ in this context is the preimage of the subset $B_i$ of $B$, and can therefore return a set (i.e., it is not a function), as will be seen presently.

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SOLUTION

(a)

Case $B_0 \subsetneq B_1$

Pick $b \in B_0$ such that the preimage in the domain $A$ is defined ($f$ may not be surjective or even one-to-one).  Apply $f^{-1}$ on such, and then $f^{-1}(b) \in f^{-1}(B_0)$.  Since $b \in B_0$, and it also belongs to $B_1$ by (proper) inclusion, $f^{-1}(b) \in f^{-1}(B_1)$.  Now pick $b \notin B_0, b \in B_1$, with defined preimage in $A$.  Apply $f^{-1}$.  Then $f^{-1}(b) \notin B_0, f^{-1}(b) \in B_1$.  Thus $f^{-1}B_0 \subsetneq f^{-1}(B_1)$.

Case $B_0 = B_1$

It is easy to see that, by arguing similarly, $f^{-1}(B_0) = f^{-1}(B_1)$.

Thus $B_0 \subset B_1 \Rightarrow f^{-1}(B_0) \subset f^{-1}(B_1)$. $\Box$

(b)

Case $f^{-1}(B_0 \cup B_1) \subset f^{-1}(B_0) \cup f^{-1}(B_1)$

Pick a $b \in B_0$, for which the preimage is deifned in $A$$f^{-1}(b) \in f^{-1}(B_0) \in f^{-1}(B_0) \cup f^{-1}(B_1)$.  Pick $b \in B_1$$f^{-1}(b) \in f^{-1}(B_1) \in f^{-1}(B_0) \cup f^{-1}(B_1)$.  We have covered every $b$ in $B_0 \cup B_1$, which preimage is $f^{-1}(B_0 \cup B_1)$.

Case $f^{-1}(B_0) \cup f^{-1}(B_1) \subset f^{-1}(B_0 \cup B_1)$

Pick a $b \in B_0$, well defined in $A$.  This element, under $f^{-1}$, belongs to $f^{-1}(B_0)$ and is therefore in $f^{-1}(B_0) \cup f^{-1}(B_1)$.  Since $b \in B_0$, it also belongs to $B_0 \cup B_1$, and has a preimage in $f^{-1}(B_0 \cup B_1)$.

Pick $b \in B_1$, well defined in $A$$f^{-1}(b) \in f^{-1}(B_1)$.  Such $b$ also belongs to $B_0 \cup B_1$, and therefore $f^{-1}(b) \in f^{-1}(B_1 \cup B_0)$.

We've now covered every $b$ in $f^{-1}(B_0) \cup f^{-1}(B_1)$, and we are done.

Thus $f^{-1}(B_0 \cup B_1) = f^{-1}(B_0) \cup f^{-1}(B_1)$. $\Box$

(c)

Case $f^{-1}(B_0 \cap B_1) \subset f^{-1}(B_0) \cap f^{-1}(B_1)$

Pick $b \in B_0 \cap B_1$ with a defined preimage in $A$.  Since it is in $B_0 \cap B_1$, the preimage lies in $f^{-1}(B_0 \cap B_1)$.

Such a $b$ belongs to $B_0$, and so its preimage is in $f^{-1}(B_0)$.  It also lies in $B_1$, and so its preimage lies also in $f^{-1}(B_1)$.  In other words, its preimage lies in $f^{-1}(B_0) \cap f^{-1}(B_1)$.

Case $f^{-1}(B_0) \cap f^{-1}(B_1) \subset f^{-1}(B_0 \cap B_1)$

Pick $b \in B_0$, with defined preimage in $f^{-1}(B_0)$.  Pick $b' \in B_1$ with defined preimage in $f^{-1}(B_1)$.  take the intersection $f^{-1}(B_0) \cap f^{-1}(B_1)$.

A $b$ mapping to the intersection has a preimage that maps both to $f^{-1}(B_0)$ and $f^{-1}(B_1)$.  In other words, such a $b \in B_0$ and $b \in B_1$, or $B_0 \cap B_1$, so the preimage of $b$ lies also in $f^{-1}(B_0 \cap B_1)$.

Thus $f^{-1}(B_0 \cap B_1) = f^{-1}(B_0) \cap f^{-1}(B_1)$. $\Box$

(d)

Case $f^{-1}(B_0 \setminus B_1) \subset f^{-1}(B_0) \setminus f^{-1}(B_1)$

Pick $b \in B_0, b\notin B_1$, with well defined preimage in $f^{-1}(B_0 \setminus B_1)$.  Since $b \in B_0$, its preimage lies in $f^{-1}(B_0)$.  But it cannot lie in $B_1$, so therefore its preimage cannot lie in $f^{-1}(B_1)$.  In other words, the preimage of such a $b$ is in $f^{-1}(B_0) \setminus f^{-1}(B_1)$.

Case $f^{-1}(B_0) \setminus f^{-1}(B_1) \subset f^{-1}(B_0 \setminus B_1)$

Pick a $b$ with preimage in $f^{-1}(B_0) \setminus f^{-1}(B_1)$.  Such a $b$ has a preimage in $f^{-1}(B_0)$, but not in $f^{-1}(B_1)$.  This means that such a $b$ must belong to $B_0$, but not $B_1$: $b \in B_0, b \notin B_1$, or, in other words, $b \in B_0 \setminus B_1$.  Such a $b$ has a preimage in $f^{-1}(B_0 \setminus B_1)$.

Thus $f^{-1}(B_0 \setminus B_1) = f^{-1}(B_0) \setminus f^{-1}(B_1)$. $\Box$

(e)

Case $A_0 \subsetneq A_1$

Pick $a \in A_0$ and apply $f$.  Such $f(a) \in f(A_0)$. By inclusion, $a \in A_1$, and therefore $f(a) \in f(A_1)$.  Pick $a \notin A_0$ but $a \in A_1$.  Apply $f$.  Then $f(a) \notin f(A_0)$ but $f(a) \in f(A_1)$.  Thus $f(A_0) \subsetneq f(A_1)$.

Case $A_0 = A_1$

Pick $a \in A_0$ (and $a \in A_1$).  Then $f(a) \in f(A_0)$ and $f(a) \in f(A_1)$.  Thus $f(A_0) = f(A_1)$.

Therefore $f(A_0) \subset f(A_1)$. $\Box$

(f)

Same as (b) but arguing form the domain side.

(g)

Case $f(A_0 \cap A_1) \subset f(A_0) \cap f(A_1)$

Pick $a \in A_0 \cap A_1$ (with image $f(a) \in f(A_0 \cap A_1)$).  Now $a \in A_0$ means that $f(a) \in f(A_0)$, and $a \in A_1$ means $f(a) \in f(A_1)$, so such an $a$'s image is in $f(A_0) \cap f(A_1)$.

Case $f(A_0) \cap f(A_1) \subset f(A_0 \cap A_1)$ if $f$ is injective

$f$ is injective means that there is a unique $a$ that yields $b \in B$, in other words $f(a) = b$.  So pick $f(a) \in f(A_0) \cap f(A_1)$.  The $a$ that was picked necessarily belongs to $A_0$ in order that $f(a)$ be in $f(A_0)$ (i.e., there isn't another $a'$ outside or inside $A_0$ whose image is $f(A_0)$, due to injectivity), but it also must belong to $A_1$ so that $f(a)$ be in $f(A_1)$ (same argument here).  Therefore $a \in A_0 \cap A_1$, and the image of this $a$ belongs to $f(A_0 \cap A_1)$.

If the argument above isn't crystal clear, the contrapositive is also insightful: if $f(A_0) \cap f(A_1) \nsubseteq f(A_0 \cap A_1)$, $f$ is not injective.  Pick $b \in f(A_0) \cap f(A_1)$ that is not also in the image $f(A_0 \cap A_1)$.  In particular this means $f^{-1}(b) = a'$ and such $a' \in A_1, a' \notin A_0$, say.  But such a $b \in f(A_0)$ and $b \in f(A_1)$, in other words, $b = f(a) \in f(A_0), f(a)\in f(A_1)$ and this means that such an $a \in A_0$ and $a \in A_1$, or $a \in A_0 \cap A_1$. The only way to resolve this apparent contradiction is by letting $b = f(a') = f(a)$, and $f$ is not injective. $\Box$

(h)

Case $f(A_0 \setminus A_1) \supset f(A_0) \setminus f(A_1)$, or $f(A_0) \setminus f(A_1) \subset f(A_0 \setminus A_1)$

Pick $f(a) \in f(A_0), f(a) \notin f(A_1)$.  Since $f(a)$ belongs to the image $f(A_0)$, there exist elements collectively $a \in A_0$, and focus on them (other elements are not of interest because they are not in $A_0$).  Since $f(a)$ does not belong to $f(A_1)$, these $a \notin A_1$. Such $a$ exactly belong to $A_0 \setminus A_1$, and therefore $f(a)$ to $f(A_0 \setminus A_1)$.

Next we want to show that if $f$ is injective, $f(A_0 \setminus A_1) \subset f(A_0) \setminus f(A_1)$.  Since all $a$s uniquely map into $f(A)$, pick any $a \in A_0 \setminus A_1$, and apply $f$.  Such will lie in $f(A_0 \setminus A_1)$.  Since $a \in A_0$, and $a \notin A_1$, this implies $f(a) \in f(A_0)$ and $f(a) \notin f(A_1)$. $\Box$

Categories:
1. May 14th, 2012 at 15:08 | #1

The statement "since $f(a)$ belongs to the image $f(A_0)$ then $a$ belongs to $A_0$" is false.

$a$ may not be an element of $A_0$ but may still be equal to $f(a)$

2. May 26th, 2012 at 15:00 | #2

Sorry, I'm not sure what you mean. You wrote: "$a$ may not be an element of $A_0$ but may still be equal to $f(a)$" makes little sense to me... I may be missing something. Perhaps what you meant is that an element outside of $A_0$ can still map to $f(a)$. If this is the case, then you are right but we are opting for those elements $a$ which definitely belong to $A_0$, because $f(a) \in f(A_0)$. I'm correcting the wording in the original post a bit to make it clearer.