## Munkres Chapter 1 Section 2

**1.**

**Problem: **Let . Let and .

**a) **Show that and that equality holds if is injective

**b) **Show that and equality holds if and only if is surjective

**Proof:**

**a) **The first part is obvious enough. If then and so from where it follows that .

Now, suppose that is injective but . Then, there is some such that . But, by definition since there exists some such that but since and it follows that and this contradicts injectivity.

Conversely, suppose that . Then, if then and so which means that as desired.

**b) **The first part is clear again. If then and so from where it follows that .

Now, suppose that is surjective. By the previous part it suffices to show the reverse inclusion. So, let (we know that any element of is of the form by surjectivity), then and so and so from where the conclusion follows.

Conversely, suppose that and let . Then, in particular from where it follows that and so surjectivity is guaranteed.

**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) ** and show that equality holds precisely when is injective

**h) ** and show that equality holds precisely when is surjective.

**Proof:**

We assume in all cases that the sets are non-empty and in the case of intersections and set differences, intersecting. For, if not this is trivial.

**a) **Let then and so and so , from where it follows that

**b) **Let then and so and so and so . Noting that all the “and so”s above were actually “if and only if”s shows the reverse inclusion.

**c) **We note that if and only if which is true if and only if which is true if and only if which is true if and only if

**d) **We note that if and only if which is true if and only if . Now, clearly the first part is true if and only if and (noticing that )the second part is true if and only if and so putting these together we get the statement in the previous sentence is true if and only if which is true if and only if

**e) **Let then for some . But, since it follows that for some and so from where it follows that .

**f) **Let then for some and thus for some and so either for some or for some and so which is true if and only if

**g) **Let then for some and so for some and for some and so and so from where it follows that .

Now, suppose that is injective. By the last part to show equality it suffices to show the reverse inclusion. To do this we let then and thus (by injectivity) and so and so . The conclusion follows by previous comment.

Conversely, suppose that then and so

from where it follows that

*Remark:* I technically took to be any subset of

**h) **Let then and and so for some but this and so for some and so

Now, suppose that is injective. From the last part to prove equality it suffices to show the reverse inclusion. So, let then (by injectivity) and so and so (second part by injectivity) and thus

Conversely, if then

from where it follows that

Injectivity follows.

*Remark:* The same remark applies.

**3.**

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

**Proof:**

We once again assume that all the following are non-empty since the proof otherwise is trivial.

**b) **Let then and so in particular for some . Thus, and so . It follows that

Conversely, let then for some . Thus, and so so that . It follows that

The problem follows.

**c) ** Let then and so for every . Thus, for every and thus

Conversely, if we have that for every and so for every . Thus, and so . It follows that

from where the problem follows.

**f) **Let then where . So, where for some . It follows that and thus from where it follows that

Conversely, let then for some . Thus, for some thus for some and so from where it follows that

and so the problem follows.

**g) **Let , then for some . It follows that for some for every . Thus, from where it follows that

Now, assume that is injective and let . Then, for every . Thus, by injectivity we have that for every and so . Thus, and thus

from where the problem follows.

**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 or ?

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

**e) **If is surjective what can you say about and ‘s surjectivity?

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

**Proof:**

**a) **Let , then and so . So, and so finally .

Conversely, if then and so so that .

The conclusion follows.

**b) **If then (by ‘s injectivity) and so (by ‘s injectivity). The conclusion follows.

**c) **We claim that is injective implies that is injective. To do this we prove the contrapositive. Suppose that was not injective, then there is such that and so clearly then and so is not injective. The conclusion follows.

**d) **This follows since as desired.

**e) **We claim that is surjective implies that is surjective. One again, we do this by proving the contrapositive. We note that from where the conclusion follows.

**f) **I’m not sure what’s desired but I guess it would be that b) and d) imply that if are bijections then is a bijection. Also, c) and e) prove that if is a bijection then is injective and surjective.

**5.**

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

Given we say the function is a *left inverse* for if . Also, we say is a *right inverse* if .

**a) **Prove that if has a left inverse, is injective, and if has a right inverse, is surjective.

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

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

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

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

**Proof:**

**a) **Suppose that has a left inverse and let . We know then that which would be impossible (since is a function$ if .

Conversely, let have right inverse and let be arbitrary. We know then that and since it is the desired value whose image is .

**b) **Consider the function . Clearly this has a left inverse but no right inverse.

**c) **How about ?

**d) **A left inverse need not be unique by this definition. For example, consider the mapping . Then, we can define by

Clearly then for every . But, we may take to be anything so that is not unique.

Right inverses are not unique either. For example, consider

We then consider two function

We then note that

and

So that

But, a similar analysis shows that if is defined by

also has the property that

And since it follows that right inverses need not be unique.

**e) **Clearly by our previous problems we have that having a left and right inverse respectively implies that $laetx f$ is injective and surjective respectively and thus bijective. Now, Let be arbitrary and suppose that

then we see by ‘s injectivity that

which is clearly absurd. Also, let be arbitrary, then by surjectivity we have that for some . Then, if this implies that

which is also a contradiction. It follows that and and thus as desired.

**6.**

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

**Proof:** We note that and so is increasing on . Thus,

but since is continuous it follows from the intermediate value theorem that and thus noting that is injective on (since it’s continuous and monotone) we may conclude that is bijective.

In problem 1 why do you feel a need to show that surjectivity and injectivity is guaranteed ?

Is not just showing “a subset of b” and “b subset of a” enough to prove

set a = set b

Comment by mehdi | May 11, 2012 |

ok I got it you said iff but the book only says if. THANKS FOR ANSWERS

Comment by mehdi | May 11, 2012 |