Article 2: Isomorphisms, Part 1

Section 3.1 Article 2: Isomorphisms, Part 1

Time to undo things! I think I’m coming into this section with a decent grasp of what an isomorphism is already, but this idea that there are two inverses instead of just one still feels a little weird to me. I find myself reminding myself that this something I already do, but now I’m just being more specific with my language.

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If teaching has taught me anything, it’s that people sometimes have very different ideas about what is meant by seemingly simple phrases like "the opposite of two". It’s often too easy to fall into a trap in assuming that since I’m thinking of two as a position on the number line and the opposite of that is defined relative to the origin, but "negative two" is really only one possible answer. If the question "what’s the oppositve of two?" had beem asked in the kitchen instead of a math class, there’s a case to be made that "one half" is actually the more appropriate response. The opposite of doubling a recipe is halving one. There might even be situations where it may be helpful to think of "opposite" as both the reciprocal and negation, as you would with slopes of perpendicular lines.

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For now, I think I need to be paying attention to the domains and codomains of the maps because that’s where I was getting tripped up earlier.

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Example 3.1.1. Exercise 1.

Show that...is an isomorphism.

Solution.

For the reflexive property (R), we can note that

A \overset{1_{A}}{\to} A

acts as it’s own inverse. Since it leaves every object in place, applying it a second time would also do nothing so

1_{A} \circ 1_{A} = 1_{A} .

If we define

f = 1_{A}

and

g = 1_{A},

then the statements

f \circ g = 1_{A}

and

g \circ f = 1_{A}

hold true and

1_{A}

is an isomorphism.

For the symmetric property (S), we’re given that

A \overset{f}{\to} B

is an isomorphism with inverse

B \overset{g}{\to} A .

By the definition of inverse, we know

g \circ f = 1_{A}

and

f \circ g = 1_{B} .

These are precisely the two statements that define

B \overset{g}{\to} A

as the inverse of

A \overset{f}{\to} B,

just with the order reversed. By including both possible composition orders in the definition, it guarantees that

f

is the inverse of

g

if and only if

g

is the inverse of

f .

For the transitive property, we’re given that

A \overset{f}{\to} B

and

B \overset{k}{\to} C

are both isomorphisms. Let

B \overset{f^{- 1}}{\to} A

be the inverse of

f

such that

f^{- 1} \circ f = 1_{A}

and

f \circ f^{- 1} = 1_{B},

and let

C \overset{k^{- 1}}{\to} B

be the inverse of

k

such that

k^{- 1} \circ k = 1_{B}

and

k \circ k^{- 1} = 1_{C} .

Consider the composition

C \overset{f^{- 1} \circ k^{- 1}}{\to} A :

Consider $(f^{- 1} \circ k^{- 1}) \circ (k \circ f) .$ By the associative property, we have $(f^{- 1} \circ k^{- 1}) \circ (k \circ f) = f^{- 1} \circ (k^{- 1} \circ k) \circ f .$ Since $k^{- 1} \circ k = 1_{B},$ it follows that $f^{- 1} \circ (k^{- 1} \circ k) \circ f = f^{- 1} \circ 1_{B} \circ f = f^{- 1} \circ f = 1_{A} .$
Consider $(k \circ f) \circ (f^{- 1} \circ k^{- 1}) .$ By the associative property, we have $(k \circ f) \circ (f^{- 1} \circ k^{- 1}) = k \circ (f \circ f^{- 1}) \circ k^{- 1} .$ Since $f \circ f^{- 1} = 1_{B},$ it follows that $k \circ (f \circ f^{- 1}) \circ k^{- 1} = k \circ 1_{B} \circ k^{- 1} = k \circ k^{- 1} = 1_{C}$
Given $(f^{- 1} \circ k^{- 1}) \circ (k \circ f) = 1_{A}$ and $(k \circ f) \circ (f^{- 1} \circ k^{- 1}) = 1_{C},$ we can then conclude that $A \overset{k \circ f}{\to} C$ is an isomorphism with an inverse of $C \overset{f^{- 1} \circ k^{- 1}}{\to} A .$

Example 3.1.2. Exercise 2.

Suppose

B \overset{g}{\to} A

and

B \overset{k}{\to} A

are both inverses...

Solution.

By definition,

B \overset{g}{\to} A

being an inverse implies

g \circ f = 1_{A}

and

f \circ g = 1_{B} .

Likewise,

B \overset{k}{\to} A

being an inverse implies

k \circ f = 1_{A}

and

f \circ k = 1_{B} .

Suppose that

g \neq k .

There would need to exist some

y \in B

with

g (y) \neq k (y) .

Denote these values

x_{1} = g (y)

and

x_{2} = k (y)

for some

x_{1}, x_{2} \in A

with

x_{1} \neq x_{2} .

Consider the point:

(k \circ f) (x_{1}) .

Since

k \circ f = 1_{A},

we must have

(k \circ f) (x_{1}) = 1_{A} (x_{1}) = x_{1} .

However, since

x_{1} = g (y)

we can also write this as

(k \circ f) (x_{1}) = (k \circ f) (g (y)) .

By the associative property,

(k \circ f) (g (y)) = k (f (g (y))) = k ((f \circ g) (y)) = k (1_{B} (y)) = k (y) = x_{2} .

Having shown that

(k \circ f) (x_{1}) = x_{1}

and

(k \circ f) (x_{1}) = x_{2},

this contradicts our earlier assumption that

x_{1} \neq x_{2}

and completes our proof that

g = k

by contradiction.

Example 3.1.3. Exercise 3.

f

has an inverse, then

f

satistfies the two cancellation laws...

Solution.

Part (a) is given in the text, so lets focus on (b). If

h \circ f = k \circ f,

it follows that

(h \circ f) \circ f^{- 1} = (k \circ f) \circ f^{- 1} .

By the associative property,

(h \circ f) \circ f^{- 1} = h \circ (f \circ f^{- 1})

and

f \circ f^{- 1} = 1_{B} .

Thus,

(h \circ f) \circ f^{- 1} = h \circ 1_{B} = h .

By similar reasoning,

(k \circ f) \circ f^{- 1} = k \circ (f \circ f^{- 1}) = k \circ 1_{B} = k .

It follows that

(h \circ f) \circ f^{- 1} = (k \circ f) \circ f^{- 1}

if and only if

h = k

which completes our proof.

Example 3.1.4. Exercise 4.

For each of the five maps above, decide whether it is invertable...

Solution.

Let’s break it down one map at time.

$R \overset{f}{\to} R$ with $f (x) = 3 x + 7$ has a well-defined inverse of $f^{- 1} (x) = (x - 7) / 3 .$
$R_{\geq 0} \overset{g}{\to} R_{\geq 0}$ with $g (x) = x^{2}$ has a well-defined inverse of $g^{- 1} (x) = \sqrt{x} .$
$R \overset{h}{\to} R$ with $h (x) = x^{2}$ does not have a well-defined inverse. We can’t ever recover the negative values after squaring them.
$R \overset{k}{\to} R_{\geq 0}$ with $k (x) = x^{2}$ also does not have a well-defined inverse. Limiting the codomain to the produced outputs doesn’t change the fact that we’re losing sign information through the map.
$R_{\geq 0} \overset{l}{\to} R_{\geq 0}$ with $l (x) = \frac{1}{x + 1}$ feels like a trick question. For most intents and purposes, we should have $l^{- 1} (x) = \frac{1}{x} - 1$ but the domain and codomain don’t look right here. Specifically, we have a problem where $l^{- 1} (0)$ is undefined and another problem where $l^{- 1} (x) < 0$ for $x > 1 .$ Based on these observations, I’d have to conclude that $l$ is not invertable.

Having still not entirely convinced myself these are correct, I thought it might be helpful to look at the graphs. I started with the first pair to make sure its what I expected.

Figure 3.1.5. Comparison of graphs for Article II, Exercise 4.1

One of the way’s I think of my inverse function is as "a reflection over the line

x = y

". This works nicely for a visual confirmation because I can tell that this relationship will hold for the rest of the lines. My confidence in

R \overset{f}{\to} R

being invertible is very high.

Now let’s take a look at the relation between

x^{2}

and

\sqrt{x} :

Figure 3.1.6. Comparison of graphs for Article II, Exercise 4.2-4.4

Notice how the inverse function only replicates half of the parabola when reflected over the diagonal. My justification for thinking that (2) is an isomorphism but (3) and (4) are not is that every point in the domain needs to have an inverse that these can be matched up with a point in the codomain (both "onto" and "one-to-one"). By including negative numbers in our domain, we’re essentially allowing un-invertable points to be passed in.

Now let’s take a look at the final functions:

Figure 3.1.7. Comparison of graphs for Article II, Exercise 4.5

On the surface, this looks like precisely the function we’re looking for. It looks like a pretty good "mirror image". The problem comes when consider the domain and codomain. The map

R_{\geq 0} \overset{l}{\to} R_{\geq 0}

is only valid for positive values, so let’s crop and zoom in our previous graph.

Figure 3.1.8. Cropped and zoomed graph for Article II, Exercise 4.5

Notice how the graph of

\frac{1}{x} - 1

seemingly has an asymptote at

x = 0

and gets "cut-off" when it hits the x-axis. For us to consider this as an inverse map, the domain of the inverse should match the codomain of the map and vice versa. Even though these maps meet my reflection criteria, the fact that

R_{\geq 0} \overset{l^{- 1}}{\to} R_{\geq 0}

is undefined for

x = 0

and

x > 1

when domain and codomain are supposed to both be

R_{\geq 0}

disqualifies it from being considered an inverse to the map

R_{\geq 0} \overset{l}{\to} R_{\geq 0} .

Perhaps what I should really be doing is looking at the compositions. We can express

l \circ l^{- 1}

\frac{1}{(\frac{1}{x} - 1) + 1}

or we can express

l^{- 1} \circ l

\frac{1}{(\frac{1}{x + 1})} - 1 .

These expressions both simplify to just

x,

but the holes left in the map when we divide by zero at

x = 0

x = - 1

don’t just disappear.

While I’m in the process of undo-ing things, I need to work on undo-ing my tendancy to rush things like this. I need to remind myself that learing this stuff isn’t a race. I’m going to use the subdivisions in Article II to break this up into smaller sections so I can give it more time to sink in.

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