## 1.2 Exercise 4

I've skipped Exercise 3 because it was basically a repeat of Exercise 2, only requiring induction.  There are plenty of induction-related proofs in the next section, but perhaps I'll return to it if the more interesting problems are depleted.

"Let $f : A \rightarrow B$ and $g : B \rightarrow C$.

(a) If $C_0 \subset C$, show that $(g \circ f)^{-1}(C_0) = f^{-1}g^{-1}(C_0)$.

(b) If $f$ and $g$ are injective, show that $g \circ f$ is injective.

(c) If $g \circ f$ is injective, what can you say about injectivity of $f$ and $g$?

(d) If $f$ and $g$ are surjective, show that $g \circ f$ is surjective.

(e)  If $g \circ f$ is surjective, what can you say about surjectivity of $f$ and $g$?

(f) Summarize your answers to (b)-(e) in the form of a theorem."

(Taken from Topology by James R. Munkres, Second Edition, Prentice Hall, NJ, 2000. Page 21.)

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SOLUTION

(a)

Pick a $c \in C_0$ with defined preimage in $A$ under $(g \circ f)^{-1}$.  In other words, a $c$ such that $(g \circ f)^{-1}(c) \in (g \circ f)^{-1}(C_0)$ (we need a defined preimage since $(g \circ f)$ may not be surjective).  Now, such a $c$ must have arrived at set $B$, and therefore must have a defined preimage in $B$, as $g^{-1}(c)$.  In turn this $g^{-1}(c)$ must arrive in $A$ via $f^{-1}$, as $(f^{-1} \circ g^{-1}) (c) \in (f^{-1} \circ g^{-1})(C_0)$.  This shows the first inclusion, that $(g \circ f)^{-1}(C_0) \subset (f^{-1} \circ g^{-1})(C_0)$.

Pick a $c$ with defined preimage in $B$ and in $A$ via $g^{-1}$ and $f^{-1}$ respectively. I.e., pick $c \in C_0$ so that $(f^{-1} \circ g^{-1})(c) \in (f^{-1} \circ g^{-1})(C_0)$.  This $c$ will have a preimage in $A$ via $(g \circ f)^{-1}$, since if it were undefined in this inverse map, it would have been undefined at either the map $g^{-1}$ or the map $f^{-1}$, but this is not the case.  In other words, $(g \circ f)^{-1}(c) \in (g \circ f)^{-1}(C_0)$.  This shows the second inclusion, that $(g \circ f)^{-1}(C_0) \supset (f^{-1} \circ g^{-1})(C_0)$.

Finally, $(g \circ f)^{-1}(C_0) = (f^{-1} \circ g^{-1})(C_0)$ $\Box$.

(b)

Suppose not (indirect proof), that $g \circ f$ is not injective.  This means that there exists $a$ and $a' \in A$ such that $(g \circ f)(a) = (g \circ f)(a')$.  But this means that we arrived at such element of $C$ by landing on the same element at $B$, meaning $f(a) = f(a')$ (which contradicts the injectivity of $f$), or by a different element of $B$, but no question at the same element after applying the map $g$, so that $g(f(a)) = g(f(a'))$.  But this contradicts the injectivity of $g$.  We must have been wrong that $(g \circ f)(a)$ is not injective, and so it must be  $\Box$.

(c)

Rephrasing the question, we want to find out  $g \circ f$ injective $\Rightarrow$ $f$ injective and, or, neither $g$ injective.  Suppose $f$ is not injective (indirect proof), and so there are elements $a$ and $a' \in A$ that map to the same element in $B$, as $f(a) = f(a')$.  Next apply $g$, so that $g(f(a)) = g(f(a'))$.  We started from two different elements in $A$, and we've now shown the map $g \circ f$ cannot be injective, a contradiction.  $f$ must therefore be injective.  Next suppose $g$ is not injective, and pick $b \neq b' \in B$, so that $g(b) = g(b')$.  If such $b, b'$ are in the image of $f$, and since $f$ is injective by the previous argument, then $b = f(a)$ and $b' = f(a')$ with $a \neq a' \in A$.  Starting from two different elements of $A$, the map $g \circ f$ cannot be injective, a contradiction. However, if, say, $b$ is not in the image of $f$, $g \circ f$ can still be one-to-one.  Thus we cannot tell that $g$ is or is not injective.

(d)

Arguing this by contradiction (indirect proof) makes the proof a very simple task.  If the map $g \circ f$ is not surjective, either $g$  is not surjective because all the elements of $B$ do not hit all elements of $C$, or if $g$ is surjective, then $f$ is not, hitting elements of $B$ for which the image under $g$ is a proper subset of $C$. (Note that those elements not in the image of $f$ in $B$ could still hit all the rest of $C$, making $g$  surjective) $\Box$.

(e)

Rephrasing, we want to find out $g \circ f$ surjective $\Rightarrow$ $f$  surjective and, or, neither $g$  surjective.  It's easiest if we start with the map $g$.  If $g$ is not surjective, there is no way $g \circ f$ can be, because we've got to stop at $B$ and take the map $g$ on our way to $C$.  Thus, $g \circ f$  surjective $\Rightarrow$  $g$ surjective. $f$  not surjective means only the image of $f$ in $B$ will map to $C$ under composition of functions. But then either such image under $g$ hits all of $C$ (no contradiction), or it doesn't (contradiction), we cannot tell.  So the behavior of $f$ is not particularly illuminating. $\Box$.

(f)

Summarizing:

$f$ and $g$ injective $\Rightarrow$  $g \circ f$ injective

$f$ and $g$ surjective $\Rightarrow$  $g \circ f$ surjective

$g \circ f$ injective $\Rightarrow$ $f$ injective

$g \circ f$ surjective $\Rightarrow$  $g$ surjective

Categories:
1. July 3rd, 2009 at 11:13 | #1

Hello. First of all, congratulations on the big task you have set up for yourself and good luck with completing it.
I am just starting to read Munkre's "Topology" and attempting to do at least some of the exercises as I go along.
I found your blog after googling for solutions to check if mine are correct.
On exercise 4, section 1.2, I don't think $g \circ f$ injective $\Rightarrow g$ injective. In your proof you say "Since $f$ is injective by the previous argument, then $b = f(a)$ and $b' = f(a')$ with $a \neq a' \in A$", but in doing so you assume that such $a$ and $a'$ exist. Since $f$ is not known to be surjective, I don't think this is necessarily true.
I'd be really happy if you could sort this out for me.

Thanks!

Manuel Araújo

2. July 3rd, 2009 at 17:41 | #2

Yes! Good catch. I made the mistake of not drawing this out! As you check your other exercises, please keep an eye out for other mistakes. I've updated the PDF files and the blog to reflect the changes. Thanks!