## Munkres Chapter one Section two: Functions

1.

**Problem:** Let . Let and

**a) **Show that and that equality holds iff

**b) **Show that and that equality holds iff is surjective

**Proof:**

**a) **Let , then$ latex f(x)\in f\left(A_0\right)$ and so . Now, suppose that is injective and let clearly then . It follows by injectivity that (if this isn’t apparent, note that by definition means that for some . Now, by injectivity we see that and so the result becomes clear). Conversely, suppose that for every . Then, in particular we see that and so

from where injectivity follows.

**b) **Let , then for some . It follows then that .

Now, suppose that is surjective and let . Then, and so . Conversely, suppose that for every . Then, in particular we see that and surjectivity follows.

2.

**Problem:** Let and let and for . Show that preserves inclusions, unions, intersections, and differences of sets:

**a) **

**b) **

**c) **

**d) **

Show that preserves inclusions and unions only:

**e) **

**f) **

**g) **; show that equality holds iff is injective

**h) **; show that equality holds iff is injective

**Proof:**

**a) **Let , then and so and thus

**b) **See number three

**c) **See number three

**d) **Let then , or it follows that and (otherwise ). This is equivalent to saying that . Conversely, let , then and . Thus, and . It follows that or

**e) **Let then for some and thus (since ) we see that for some which is equivalent to saying that .

**f) **See number three

**g) **See number three for the first part. Clearly, it suffices to prove the second part for two sets. So, suppose that is injective and let then and so and and so and , or . The proposed equality follows.

Conversely, suppose that for every . We see then in particular the following line of reasoning

and thus

and clearly singletons are disjoint iff they’re single elements are not equal. Thus, condensing

and injectivity follows.

**h) **Let , then and . This, in particular says that for some and (otherwise . Thus, for some , or .

Now, suppose that is injective and let , then by injectivity and so and . Thus, and . So, . Conversely, if then we see that

but iff . It follows that and so .

3.

**Problem:** Show that b), c), f), and g) of the last exercise hold for arbitrary unions and intersections

**Proof:**

**a) **Let then . Thus, there exists some such that and so . Therefore, .

Conversely, if let , then for some . Therefore, so that . So finally, we may conclude that

**c) **We merely need note that

(where the first equality is gotten noticing that and using property d) from the previous problem). Then, recalling the last problem we can see that this is equal to

and so comparing the LHS of the first equality with the RHS of the last leads to the desired result.

**f) **Let . Then, for some . Thus, for some for some . Thus, and so .

Conversely, let . Then, for some . It follows that for some and so for some . Therefore,

**g)** Let . Then, for some . It follows that for some for every . Thus, for every and so

4.

**Problem: **Let and

**a) **If , show that

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

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

**d) **If and are surjective, prove that is surjective

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

**Proof:**

**a) **Let , then but this clearly implies that . Conversely, if we see that and so . Therefore,

**b) **If then by the injectivity of we see that and so by the injectivity of it follows that .

**c) **If is injective then is injective. To see this, suppose not; then there exists such that but which contradicts ‘s injectivity.

**d) **We note that from where the conclusion follows.

**e) **It must be true that is surjective. To see this suppose not. Then,

which contradicts the surjectivity of

5.

**Problem: **In general, let us denote the identity function on a set by . That is, define

Given we say that a function is a left inverse for if ; and we see that is a right inverse for if .

**a) **Show that if has a left inverse, is injective; and if has a right inverse, is surjective

**b) **Give an example of a function that has a left inverse but no right inverse

**c) **Give an example of a function which has a right inverse but no left inverse.

**d) **Can a function have more than one left inverse? More than one right inverse?

**e) **Show that if has both a left inverse and a right inverse , then is bijective and

**Proof:**

**a) **If we see that and so is injective. Conversely, let be arbitrary, we know that and so is the required element which maps to under

**b) **The inclusion map . Clearly it has no right inverse or it’d be surjective. But, the mapping

surely satisfies the left inverse requirement since

.

**c) **Consider the function . Clearly this possesses no left inverse, otherwise it’d be injective. But, has the quality that and so is a suitable right inverse

**d) **Yes, in the first example we could have taken and the second example we could have had , yet both are left and right inverses respectively.

**e) **Clearly by a) and b) we see that is bijective. Furthermore, let be arbitrary (we can say since it’s surjective), then

and

where is furnished by ‘s injectivity. Thus, putting the two together gives .

6.

**Problem: **Let . By restricting the domain and range of obtain from a bijective function .

**Proof:** Merely take and and so then the function is bijective and a restriction of (in fact, you could take the vaccuous function )

Wow! That looks impressive. What prerequisites should one meet to understand this stuff (how much of set theory etc)?

Comment by Student | September 25, 2010 |

I can’t tell if you’re a real person or a bot haha. Though, your e-mail is hoffmankunze which are the authors of a good lin. alg. book. To answer your question: this stuff really isn’t that hard and the only prerequisites are the stuff in the prior post (Munkres section one). This is that set theory.

Comment by drexel28 | September 25, 2010 |

Definitely not a bot. As you have deducted, it would have been a miracle if a bot could appreciate the good book so as to turn the authors into its namesake. 😀

That’s a relief for me then. Someone told me that it’s futile to even try learning topology if one hasn’t got a complete mastery of set theory, so I was too afraid to even look at a topology book. All the set theory I know was learnt from the preliminary section of Herstein’s Topics In Algebra, so it isn’t strong. (I need to learn some set theory, anyway, since it seems to be everywhere; so let me know if you know any good books, please). Thanks.

Comment by Student | September 26, 2010

I’m sorry friend! It’s true that topology is probably one of the most heavily set theoretic subjects commonly encountered…besides set theory itself. If you’ve done Herstein (or algebra in general, though Herstein is a very good book) you should be well acquainted with the function aspect of set theory. If you want to learn topology you can pick up Munkres and just start. It’s fairly well self-contained and teaches you all the set theory you need to know (in fact, it will teach you as much or more set theory than most undergraduates/graduates know). If you’re looking for something more comprehensive I LOVED the book Set Theory and Logic by Stoll. If found it to be a comprehensive (and abstract 🙂 ), but readable, exposition on all the set theory a non-set theorist (or set-theoretic topologist) could want to know. In fact, he even goes into some serious logic including the infamous Goedel’s theorem.

If you have any questions whatsoever, feel free to contact me on here or at afy25@drexel.edu

Comment by drexel28 | September 26, 2010 |

Thanks. I will get both books. The latter merely because I want this burden of set theory that seems to be everywhere not to bother me ever again. But I’m glad that Munkres contains all the set theory that one would need to go through the book and more. That really makes it a brilliant book. Yes, I’ve done Herstein but not yet any other book on Algebra (counting Linear Algebra out – I’ve done Hoffman and Kunze, as you have gathered). I will contact you to let you know how it goes. You are very kind. Thank you.

Comment by Student | September 27, 2010 |

No problem friend, anything for a fellow math enthusiast!

Comment by drexel28 | September 27, 2010 |

I am Industrial engineer and I am studying the munkres book by my own. I think this have wonderful applications in my field but at the beginning they are difficult to see. If you know about that kind of stuff I would be happy to talk about.

Thanks a lot for sharing your work

Comment by Luis Felipe Cardona | July 18, 2011 |