## 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 and .

(a) If , show that .

(b) If and are injective, show that is injective.

(c) If is injective, what can you say about injectivity of and ?

(d) If and are surjective, show that is surjective.

(e) If is surjective, what can you say about surjectivity of and ?

(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 with defined preimage in under . In other words, a such that (we need a defined preimage since may not be surjective). Now, such a must have arrived at set , and therefore must have a defined preimage in , as . In turn this must arrive in via , as . This shows the first inclusion, that .

Pick a with defined preimage in and in via and respectively. I.e., pick so that . This will have a preimage in via , since if it were undefined in this inverse map, it would have been undefined at either the map or the map , but this is not the case. In other words, . This shows the second inclusion, that .

Finally, .

(b)

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

(c)

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

(d)

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

(e)

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

(f)

Summarizing:

and injective injective

and surjective surjective

injective injective

surjective surjective

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 injective injective. In your proof you say "Since is injective by the previous argument, then and with ", but in doing so you assume that such and exist. Since 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

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!