# Exercises:Saul - Algebraic Topology - 1

## Exercises

### Exercise 1.2

Arrange the capital letters of the Roman alphabet thought of as graphs into homeomorphism classes.

#### Solutions

First note I will use the font provided by \sf, giving the following letters: [ilmath]\sf{ABCDEFGHIJKLMNOPQRSTUVWXYZ} [/ilmath]. I notice that [ilmath]\sf{G} [/ilmath] here is homeomorphic to a [ilmath]C[/ilmath], so I have included [ilmath]\underline{\text{G} } [/ilmath], this represents [ilmath]G[/ilmath] with a [ilmath]\top[/ilmath] shape where the [ilmath]\sf{G} [/ilmath] only has a [ilmath]\rceil[/ilmath]. I've done both because the [ilmath]\top[/ilmath]-like form is so common it is worth doing and apparently a sans-serif font lacks this "T" part of the G.

• I also include [ilmath]\mathcal{Z} [/ilmath] representing a [ilmath]\sf{Z} [/ilmath] with a - through the middle, again due to how common this form is
• I also include [ilmath]\text{I} [/ilmath], as this is also really common for a capital "I", along side [ilmath]\sf{I} [/ilmath], the font's version of a capital "I"
• I also include [ilmath]\text{J} [/ilmath], as this is again really common, and how I write them.
• I also include [ilmath]\sf{k} [/ilmath], as it's very common to see them as a [ilmath]<[/ilmath] joined at the point to a [ilmath]\vert[/ilmath]

The homeomorphisms claimed to exist and described here in words should all be shown in an appendix to this print out (or in a picture below for online readers) showing, in pictorial form, what is intended. There will also be pictures of the additional letters represented by [ilmath]\underline{\text{G} } [/ilmath], [ilmath]\sf{K} [/ilmath] and [ilmath]\mathcal{Z} [/ilmath].

The important lemmas and theorems can be found below the table at the bottom of this document.

The homeomorphism classes are:

• [ilmath]\{\sf{A, R}\} [/ilmath]
• [ilmath]\{\sf{B} \} [/ilmath]
• [ilmath]\{\sf{C, G, I, J, L, M, N, S, U, V, W, Z} \} [/ilmath]
• [ilmath]\{\sf{D, O} \} [/ilmath]
• [ilmath]\{\sf{E, F, } [/ilmath][ilmath]\underline{\text{G} } [/ilmath][ilmath]\sf{, } [/ilmath][ilmath]\text{J} [/ilmath][ilmath]\sf{, T, Y} \} [/ilmath]
• [ilmath]\{\sf{H, } [/ilmath][ilmath]\text{I} [/ilmath][ilmath]\sf{, K} \} [/ilmath]
• [ilmath]\{\sf{P} \} [/ilmath]
• [ilmath]\{\sf{Q} \} [/ilmath]
• [ilmath]\{\sf{k, X, } [/ilmath][ilmath]\mathcal{Z}\} [/ilmath]
##### Reasoning
Letter Class so far Reasoning Comment
[ilmath]\sf{A} [/ilmath] [ilmath]\{\sf{A}\} [/ilmath] There are no classes yet. So [ilmath]\sf{A} [/ilmath] founds one
[ilmath]\sf{B} [/ilmath] [ilmath]\{\sf{B}\} [/ilmath] Crafty[Note 1] point removal[Note 2] - by removing a point from the bottom right (\) edge of the [ilmath]\sf{A} [/ilmath] or the bottom left (/) edge we end up with 2 pathwise connected components (herein: components). However removing any point from [ilmath]\sf{B} [/ilmath] results in one component.
[ilmath]\sf{C} [/ilmath] [ilmath]\{\sf{C}\} [/ilmath] There are no loops in [ilmath]\sf{C} [/ilmath] (it is obviously homeomorphic to just a line ([ilmath]\vert[/ilmath]) say, due to the absence of holes (of which [ilmath]\sf{A} [/ilmath] has one and [ilmath]\sf{B} [/ilmath] has two - see fundamental group) we must conclude [ilmath]\sf{C} [/ilmath] is none of the existing groups and founds its own.
[ilmath]\sf{D} [/ilmath] [ilmath]\{\sf{D}\} [/ilmath]
• By Crafty point removal we see removing any point from [ilmath]\sf{D} [/ilmath] leaves one component, where as a crafty choice of point can leave [ilmath]\sf{A} [/ilmath] with 2 components.
• By noticing the fundamental group of [ilmath]\sf{B} [/ilmath] would be [ilmath]\mathbb{Z}*\mathbb{Z} [/ilmath] and the fundamental group of [ilmath]\sf{D} [/ilmath] will be that of the circle, [ilmath]\mathbb{Z} [/ilmath] we see that [ilmath]\sf{D} [/ilmath] is not homeomorphic to [ilmath]\sf{B} [/ilmath]
[ilmath]\sf{E} [/ilmath] [ilmath]\{\sf{E}\} [/ilmath] By Crafty point removal we can have 3 components. No member of any class so far has this property. Therefore [ilmath]\sf{E} [/ilmath] must have its own class
[ilmath]\sf{F} [/ilmath] [ilmath]\{\sf{E, F}\} [/ilmath] A continuous map that doubles the length of the bottom [ilmath]\vert[/ilmath] of the [ilmath]F[/ilmath] and bends the latter half of it at a right angle to the right is easily seen to be an [ilmath]\sf{E} [/ilmath] and the inverse map simply shortens the [ilmath]\lfloor[/ilmath]-like part of the [ilmath]\sf{E} [/ilmath] and "unkinks" the right angle.
[ilmath]\sf{G} [/ilmath] [ilmath]\{\sf{C, G}\} [/ilmath] If you "retract" the [ilmath]-[/ilmath] part of the [ilmath]\sf{G} [/ilmath] and shorten the resulting [ilmath]\sf{C} [/ilmath] like shape until it is a [ilmath]\sf{C} [/ilmath] - clearly the inverse of this map involves extending the bottom arc of a [ilmath]\sf{C} [/ilmath] then bending it to a right angle is also continuous, thus homeomorphism. [ilmath]\top[/ilmath] [ilmath]\sf{G} [/ilmath]
[ilmath]\underline{\text{G} } [/ilmath] [ilmath]\{\sf{E, F, } [/ilmath][ilmath]\underline{\text{G} } [/ilmath][ilmath]\} [/ilmath] Remembering that a [ilmath]\underline{\text{G} } [/ilmath] is supposed to represent a [ilmath]\sf{G} [/ilmath] with a [ilmath]\top[/ilmath] rather than a [ilmath]\rceil[/ilmath] bit in the middle, we note that it is homeomorphic to a [ilmath]\sf{T} [/ilmath], if you then take the tail (the [ilmath]\vert[/ilmath]) of the [ilmath]\sf{T} [/ilmath] and wrap it around in a [ilmath]C[/ilmath] shape and we have what we represent by [ilmath]\underline{\text{G} } [/ilmath], this transformation of [ilmath]\sf{T} [/ilmath] into [ilmath]\underline{\text{G} } [/ilmath] is obviously a homeomorphism [ilmath]\text{G} [/ilmath] - shown as it is similar to the G being described.
[ilmath]\sf{H} [/ilmath] [ilmath]\{\sf{H}\} [/ilmath]
• We can rule out [ilmath]\{\sf{A}\} [/ilmath], [ilmath]\{\sf{B}\} [/ilmath] and [ilmath]\{\sf{D}\} [/ilmath] by noting the absence of holes of [ilmath]\sf{H} [/ilmath]
• We can rule out [ilmath]\{\sf{C,\ldots}\} [/ilmath] by Crafty point removal, notice that removing any point from [ilmath]\sf{C} [/ilmath] results in one (the tips of the arc) or two (anywhere else) components. Yet removing one point from [ilmath]\sf{H} [/ilmath] can yield 3 components. (For example the intersection of the horizontal and vertical parts of the [ilmath]\vdash[/ilmath] part of [ilmath]\sf{H} [/ilmath])
• We can rule out [ilmath]\{\sf{E,\ldots}\} [/ilmath] by Two-step point removal, suppose [ilmath]{\sf{H} }\cong{\sf E} [/ilmath], pick the two points to be where the horizontal - of the [ilmath]\sf{H} [/ilmath] meets the vertical sides. Then we have 5 components. However removing any two points from [ilmath]\sf{E} [/ilmath] leaves us with at most 4 components. Contradicting that they're homeomorphic.

Thus we give [ilmath]\sf{H} [/ilmath] its own class

[ilmath]\sf{I} [/ilmath] [ilmath]\{\sf{C, G, I}\} [/ilmath] Trivial - just take the [ilmath]\sf{I} [/ilmath] and bend it slightly, obviously reversible and continuous each way, therefore homeomorphism
[ilmath]\text{I} [/ilmath] [ilmath]\{\sf{H, } [/ilmath][ilmath]\text{I}\} [/ilmath] Trivial - just take the [ilmath]\text{I} [/ilmath] rotate it and stretch it a little into shape so it's a [ilmath]\sf{H} [/ilmath], this is obviously continuous, as is the inverse of contracting the "height" of the [ilmath]\sf{H} [/ilmath] then rotating it, so it's an [ilmath]\sf{I} [/ilmath]
[ilmath]\sf{J} [/ilmath] [ilmath]\{\sf{C, G, I, J }\} [/ilmath] Obviously homeomorphic to [ilmath]\sf{I} [/ilmath] as if you take the [ilmath]\sf{I} [/ilmath] and bend the bottom round you have a [ilmath]\sf{J} [/ilmath], obviously continuous, as is the inverse of "straightening out" the [ilmath]\sf{J} [/ilmath] to yield [ilmath]\sf{I} [/ilmath]
[ilmath]\text{J} [/ilmath] [ilmath]\{\sf{E, F, } [/ilmath][ilmath]\underline{\text{G} } [/ilmath][ilmath]\sf{, } [/ilmath][ilmath]\text{J} [/ilmath][ilmath]\} [/ilmath] We shall see later that a [ilmath]\sf{T} [/ilmath] is homeomorphic to an [ilmath]\sf{E} [/ilmath] (by rotation and extending the branches and bending them to be parallel to the "trunk"). I claim now that [ilmath]\sf{T} [/ilmath] is homeomorphic to [ilmath]\text{J} [/ilmath], we do this by extending the "serifs" at the top of the [ilmath]\text{J} [/ilmath] to form the branches at the top of the [ilmath]\sf{T} [/ilmath] and meanwhile we shorten and straighten out the curved bottom part of the [ilmath]\text{J} [/ilmath] (like we did with [ilmath]\sf{I} [/ilmath] and [ilmath]\sf{J} [/ilmath] above). Clearly this is continuous, and its inverse - of shrinking the branches of the [ilmath]\sf{T} [/ilmath], lengthening and bending round the "trunk" to form a [ilmath]\text{J} [/ilmath] - is also continuous. As homeomorphism is an equivalence relation we see [ilmath]\text{J}\cong [/ilmath][ilmath]\sf{E} [/ilmath], as required.
[ilmath]\sf{K} [/ilmath] [ilmath]\{\sf{H, I, K}\} [/ilmath] Upon careful inspect we see that the the [ilmath]\sf{K} [/ilmath] is really a [ilmath]\sf{Y} [/ilmath] rotated [ilmath]180^\circ[/ilmath] joined onto a [ilmath]\vert[/ilmath]. This is easily seen to be homeomorphic to a [ilmath]\sf{H} [/ilmath] by a little rotating of the edges of the [ilmath]\sf{Y} [/ilmath] part and a little stretching, as usual, the inverse of this map is also obviously continuous.
[ilmath]\sf{k} [/ilmath] [ilmath]\{\sf{k}\} [/ilmath] Removing the point where the [ilmath]<[/ilmath] meets the [ilmath]\vert[/ilmath] we have 4 components. No other letter done so far has four components with one point removed (highest is 3) - therefore must have a new class made.
[ilmath]\sf{L} [/ilmath] [ilmath]\{\sf{C, G, I, J, L}\} [/ilmath] If you take the [ilmath]\sf{I} [/ilmath], extend it, then put a right-angled kink in the bottom we obtain an [ilmath]\sf{L} [/ilmath], this is clearly continuous, the inverse, which straightens out then shortens is also easily seen to be continuous, thus homeomorphic.
[ilmath]\sf{M} [/ilmath] [ilmath]\{\sf{C, G, I, J, L, M}\} [/ilmath] Take the [ilmath]I[/ilmath], rotate it so its horizontal, stretch it, then put some kinks in it to make the [ilmath]\sf{M} [/ilmath] shape, obviously continuous, the inverse of unkinking, shortening, then rotating so its vertical again is also easily seen to be continuous, thus we have a homeomorphism
[ilmath]\sf{N} [/ilmath] [ilmath]\{\sf{C, G, I, J, L, M, N}\} [/ilmath] Take the [ilmath]I[/ilmath], rotate it so its horizontal, stretch it, then put some kinks in it to make the [ilmath]\sf{N} [/ilmath] shape, obviously continuous, the inverse of unkinking, shortening, then rotating so its vertical again is also easily seen to be continuous, thus we have a homeomorphism
[ilmath]\sf{O} [/ilmath] [ilmath]\{\sf{D, O}\} [/ilmath] Take the [ilmath]\mathsf{D} [/ilmath] and curve the straight edge slightly until it becomes an [ilmath]\sf{O} [/ilmath] (they look to be the same size, but any "nudging" around to make it identical to the [ilmath]\sf{O} [/ilmath] is a homeomorphism itself), the inverse of the nudging homeomorphism (if required) followed by the straightening that curved edge yields a [ilmath]\sf{D} [/ilmath] and is also obviously continuous, thus a homeomorphism
[ilmath]\sf{P} [/ilmath] [ilmath]\{\sf{P}\} [/ilmath] [ilmath]\sf{P} [/ilmath] clearly has a hole, so we can rule out all the classes of hole-less letters. It only has one hole, so we can rule out [ilmath]\{\sf{B}\} [/ilmath]. We're left with the [ilmath]\{\sf{A}\} [/ilmath] and the [ilmath]\{\sf{D, O}\} [/ilmath] classes.
• Take [ilmath]\sf{O} [/ilmath] and remove one point, we are left with one component. Take [ilmath]\sf{P} [/ilmath] and remove any point (except the end) of the "stem" coming off the "o" part. We now have 2 components. So these cannot be homeomorphic.
• Take [ilmath]\sf{A} [/ilmath], suppose it is homeomorphic (via [ilmath]f[/ilmath]) to [ilmath]\sf{P} [/ilmath], then take [ilmath]p\in{\sf A} [/ilmath] to be the point where the [ilmath]/[/ilmath] part means the [ilmath]-[/ilmath] part; this splits [ilmath]\sf{A} [/ilmath] into two components. By hypothesis [ilmath]\mathcal{O}(f(p))[/ilmath] must also be 2, so pick [ilmath]f(p)[/ilmath] to be anywhere on the stem on the [ilmath]\sf{P} [/ilmath] - up to and including where the loop and the stem meet, but not the end of the stem, then that splits [ilmath]\sf{P} [/ilmath] into two components also.
• Indeed any other point would split [ilmath]\sf{P} [/ilmath] into only one connected component
• Now delete the point, say [ilmath]q[/ilmath] where the [ilmath]\backslash[/ilmath] meets the [ilmath]-[/ilmath]. [ilmath]\sf{A} [/ilmath] now has 4 connected components.
• However for all [ilmath]f(q)\in {\sf P}-\{f(p)\} [/ilmath] we only have 2, or at most 3 path components. For none of these points can we have 4
• This contradicts that [ilmath]\sf{A} [/ilmath] and [ilmath]\sf{P} [/ilmath] are homeomorphic.
• So [ilmath]\sf{P} [/ilmath] does not belong in [ilmath]\{\sf{A} \} [/ilmath]

Thus we give [ilmath]\sf{P} [/ilmath] its own class

[ilmath]\sf{Q} [/ilmath] [ilmath]\{\sf{Q}\} [/ilmath] [ilmath]\sf{Q} [/ilmath] has one loop, this leaves us with [ilmath]\{\sf{A}\} [/ilmath] or [ilmath]\{\sf{D, O}\} [/ilmath] as potential candidates ([ilmath]\{\sf{B}\} [/ilmath] is ruled out as it has two holes)
• Remove the part where the [ilmath]\circ[/ilmath] meets the [ilmath]\backslash[/ilmath] of the [ilmath]\sf{Q} [/ilmath] and we are left with 3 components.
• Remove any point from [ilmath]\sf{A} [/ilmath] and we are left with 1 or 2 components - so [ilmath]\sf{A} [/ilmath] cannot be homeomorphic to [ilmath]\sf{Q} [/ilmath]
• Remove any point from [ilmath]\sf{O} [/ilmath] and we are left with 1 component - so [ilmath]\sf{O} [/ilmath] cannot be homeomorphic to [ilmath]\sf{Q} [/ilmath]
[ilmath]\sf{R} [/ilmath] [ilmath]\{\sf{A, R}\} [/ilmath] It is easy to see that with minimal distorting we can construct a homeomorphism between [ilmath]\sf{A} [/ilmath] and [ilmath]\sf{R} [/ilmath]. Truly.
[ilmath]\sf{S} [/ilmath] [ilmath]\{\sf{C, G, I, J, L, M, N, S}\} [/ilmath] It is easy to see that [ilmath]\sf{S} [/ilmath] and [ilmath]\sf{I} [/ilmath] are homeomorphic. Simply take the [ilmath]\sf{I} [/ilmath] and start to curve the top half to the left and the bottom half to the right, one will end up with an [ilmath]\sf{S} [/ilmath]. For the inverse it will just be taking an [ilmath]\sf{S} [/ilmath] and pushing those curves towards the centre to make an [ilmath]\sf{I} [/ilmath] - this is clearly a homeomorphism
[ilmath]\sf{T} [/ilmath] [ilmath]\{\sf{E, F, } [/ilmath][ilmath]\underline{\text{G} } [/ilmath][ilmath]\sf{, } [/ilmath][ilmath]\text{J} [/ilmath][ilmath]\sf{, T} \} [/ilmath] Although the [ilmath]\text{J} [/ilmath] and the [ilmath]\sf{T} [/ilmath] would probably be easier to show, I must show [ilmath]\sf{E} [/ilmath] is homeomorphic to [ilmath]\sf{T} [/ilmath] as earlier work depends upon it. Rotate [ilmath]\sf{T} [/ilmath] [ilmath]90^\circ[/ilmath] anti-clockwise and then extend the branches of the [ilmath]\sf{T} [/ilmath] until they are long enough to be kinked at a right angle towards the trunk of the [ilmath]\sf{T} [/ilmath], yielding an [ilmath]\sf{E} [/ilmath]. This is obviously a homeomorphism as each step is bijective, continuous and has continuous inverse.
[ilmath]\sf{U} [/ilmath] [ilmath]\{\sf{C, G, I, J, L, M, N, S, U}\} [/ilmath] Clearly homeomorphic to [ilmath]sf{C} [/ilmath] as if you take a [ilmath]\sf{C} [/ilmath] and rotate it [ilmath]90^\circ[/ilmath] anti-clockwise, then make it a bit thinner, and stretch the prongs so they're longer, you end up with a [ilmath]\sf{U} [/ilmath], each step is obviously continuous and bijective with continuous inverse, so the composition is a homeomorphism
[ilmath]\sf{V} [/ilmath] [ilmath]\{\sf{C, G, I, J, L, M, N, S, U, V}\} [/ilmath] This is another simple homeomorphism, [ilmath]\sf{U} [/ilmath] is homeomorphic to [ilmath]\sf{V} [/ilmath] simply by straightening out each half (split by a vertical axis going through the minimum of the [ilmath]\sf{U} [/ilmath]), the result will be [ilmath]\sf{V} [/ilmath], this is clearly bijective, continuous, and "curving" a [ilmath]\sf{V} [/ilmath] (at the lower 1/3rd or so of the arms) - the inverse - is clearly continuous.
[ilmath]\sf{W} [/ilmath] [ilmath]\{\sf{C, G, I, J, L, M, N, S, U, V, W}\} [/ilmath] [ilmath]\sf{W} [/ilmath] is clearly homeomorphic to [ilmath]\sf{M} [/ilmath], we simply need to rotate it [ilmath]180^\circ[/ilmath], then widen out the end points of the arms (whilst keeping them straight) to yield a [ilmath]\sf{W} [/ilmath].
• We could also take a [ilmath]V[/ilmath], fix the upper halfs of each component if the tip of the [ilmath]\vee[/ilmath] part were removed, then pull that apex up until we get a [ilmath]\sf{W} [/ilmath] shape, then enlarge as needed, but this is harder to describe using words.
[ilmath]\sf{X} [/ilmath] [ilmath]\{\sf{k, X}\} [/ilmath] We consider [ilmath]\sf{k} [/ilmath] as 4 edges joined at a vertex. [ilmath]\sf{X} [/ilmath] is the same thing. We just have to rotate the edges in a manner similar to the hands on a clock. These rotations about a point are obviously homeomorphisms, job done.
[ilmath]\sf{Y} [/ilmath] [ilmath]\{\sf{E, F, } [/ilmath][ilmath]\underline{\text{G} } [/ilmath][ilmath]\sf{, } [/ilmath][ilmath]\text{J} [/ilmath][ilmath]\sf{, T, Y} \} [/ilmath] [ilmath]\sf{Y} [/ilmath] is homeomorphic to [ilmath]\sf{T} [/ilmath], to see this simply take the [ilmath]\sf{T} [/ilmath] and move the vertex where the 3 edges meet down until one has a [ilmath]\sf{Y} [/ilmath] shape, this is clearly continuous, bijective and the reverse is continuous also, thus a homeomorphism.
[ilmath]\sf{Z} [/ilmath] [ilmath]\{\sf{C, G, I, J, L, M, N, S, U, V, W, Z}\} [/ilmath] [ilmath]\sf{Z} [/ilmath] and [ilmath]\sf{N} [/ilmath] are easily seen to be homeomorphic, simply rotate [ilmath]\sf{N} [/ilmath] clockwise [ilmath]90^\circ[/ilmath], this is a homeomorphism. Some vertical stretching may also be in order, but that is also a homeomorphism.
[ilmath]\mathcal{Z} [/ilmath] [ilmath]\{\sf{k, X, } [/ilmath][ilmath]\mathcal{Z} [/ilmath][ilmath]\sf{}\} [/ilmath] We will show that [ilmath]\mathcal{Z} [/ilmath] (intended to represent a [ilmath]\sf{Z} [/ilmath] with a [ilmath]-[/ilmath] through the [ilmath]/[/ilmath] part) is homemorphic to [ilmath]\sf{X} [/ilmath]. To do so take an [ilmath]\sf{X} [/ilmath] and stretch the top right spoke to roughly double its length and kink it off to the left at roughly half way. We have now formed the top part of the [ilmath]\mathcal{Z} [/ilmath]. Next we take the bottom left spoke and perform a similar operation, extending it to approximately double its length, then kinking it off to the right this time, we have now formed the [ilmath]\sf{Z} [/ilmath] of the [ilmath]\mathcal{Z} [/ilmath] shape. We now take the top left and bottom right "spokes" of the [ilmath]\sf{X} [/ilmath] we started with and - whilst keeping them parallel - rotate them about the central point until they are horizontal. Then we shrink as needed to form the [ilmath]\mathcal{Z} [/ilmath] shape. This is clearly continuous, as is its inverse, and bijective. Thus a homeomorphism. [ilmath]\mathcal{Z} [/ilmath]

#### Lemmas

Caveat:The following listed here are for reference to someone looking at the exercises only. They are done from memory and have no reference (at the time of writing) - use at your own peril.

##### Homeomorphisms and point removal

Suppose [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath] are topological spaces and [ilmath]f:X\rightarrow Y[/ilmath] is a homeomorphism between them (so [ilmath]X\cong Y[/ilmath]), then for any [ilmath]x\in X[/ilmath] we have:

##### Point removal

Let [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath] be topological spaces and let [ilmath]f:X\rightarrow Y[/ilmath] be any map (possibly continuous) between them. Then

• [ilmath]f:X\rightarrow Y[/ilmath] being a homeomorphism implies [ilmath]\forall p\in X[\mathcal{O}(p)\eq\mathcal{O}(f(p))][/ilmath] - where [ilmath]\mathcal{O}(p)[/ilmath] is the number of path-connected components of the space [ilmath]X-\{p\} [/ilmath]
• We may write [ilmath]\mathcal{O}(p,X)[/ilmath] to mean [ilmath]p[/ilmath] removed from the space [ilmath]X[/ilmath], this makes things clearer when dealing with subsets.

Specifically, by contrapositive, if [ilmath]\exists p\in X[\mathcal{O}(p)\neq\mathcal{O}(f(p))][/ilmath] then [ilmath]f:X\rightarrow Y[/ilmath] is not a homeomorphism.

##### Two-step point removal

This is a corollary of the two claims above. Two-step point removal means that:

• Suppose [ilmath]X\cong_f Y[/ilmath], then [ilmath]X-\{p\}\cong_{f\vert_{X-\{x\} } }Y-\{f(p)\} [/ilmath] (by "homeomorphisms and point removal")
• Suppose [ilmath]X-\{p\}\cong_{f\vert_{X-\{x\} } }Y-\{f(p)\} [/ilmath] are indeed homeomorphic, then [ilmath]\forall q\in X[\mathcal{O}(q,X-\{p\})\eq\mathcal{O}(f\vert_{X-\{x\} }(q);Y-\{f(p)\})][/ilmath]

We conclude:

• [ilmath]X\cong_f Y[/ilmath] implies [ilmath]\forall q\in X[\mathcal{O}(q,X-\{p\})\eq\mathcal{O}(f\vert_{X-\{x\} }(q);Y-\{f(p)\})][/ilmath]

### Exercise 1.5

 The matrix of [ilmath]A:\mathbb{Z}^3\rightarrow\mathbb{Z}^3[/ilmath] $A:\eq\left(\begin{array}{ccc}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7& 8 &9\end{array}\right)$
Let [ilmath]A:\mathbb{Z}^3\rightarrow\mathbb{Z}^3[/ilmath] be the group homomorphism given by the matrix on the right. Calculate its kernel, image and cokernel.
• Alec's note:
• [ilmath]\text{Image}(A):\eq A(\mathbb{Z}^3):\eq\{y\in\mathbb{Z}^3\ \vert\ \exists x\in\mathbb{Z}^3[A(x)\eq y]\}\eq\{A(x)\ \vert\ x\in\mathbb{Z}^3\} [/ilmath]
• [ilmath]\text{Ker}(A):\eq\{x\in\mathbb{Z}^3 \ \vert\ A(x)\eq 0\} [/ilmath]
• TODO: Cokernel is the homology group here. Check that

#### Solution

##### Image of [ilmath]A[/ilmath]

It is easy to see Caveat:In linear algebra.... over vector spaces.... that the span of the images of basis elements spans the image. We will use this here.

• [ilmath]\text{Im}(A)\eq\text{Span}\left(\left\{A\left(\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)\right),A\left(\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)\right),A\left(\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\right\}\right)\right)[/ilmath] [ilmath]\eq\text{Span}\left(\left\{\left(\begin{array}{c}1\\ 4\\ 7\end{array}\right),\left(\begin{array}{c}2\\ 5\\ 8\end{array}\right),\left(\begin{array}{c}3\\ 6\\ 9\end{array}\right)\right\}\right)[/ilmath]

Let us now reduce this to a basis (and write these as row-vectors transposed from now on)

• Clearly [ilmath](1,4,7)^T[/ilmath] and [ilmath](2,5,8)^T[/ilmath] are "linearly independent"
• To see if the 3rd vector is linearly independent of these two we must
• Image - the first 2 columns of the matrix form a vector. RREFfing [ilmath]A[/ilmath] shows that -1 x the first vector + 2 x the second vector = the third vector
• Kernel - RREFfing [ilmath]A[/ilmath] with an extra column of zeros shows the kernel has basis [ilmath](1,-2,1)[/ilmath] or something.

Caveat:This is "dumb luck", it could be nastier, there's no.... like all of these are isomorphic to copies of [ilmath]\mathbb{Z} [/ilmath]

• Co-kernel, best thought of as $\frac{\langle (1,0,0)^T,(0,1,0)^T,(0,0,1)^T\rangle}{\langle (1,4,7)^T,(2,5,8)^T\rangle }$
• Clearly we can replace [ilmath](1,0,0)^T[/ilmath] with [ilmath](1,4,7)^T[/ilmath] as [ilmath]a(1,4,7)-4a(0,1,0)-7a(0,0,1)\eq a(1,0,0)[/ilmath]
• We cannot do the same thing again, as [ilmath]a(1,4,7)+b(2,5,8)+c(0,0,1)[/ilmath] can never be [ilmath](0,1,0)[/ilmath] (RREFing we get fractional values. Which is telling...)
###### Thoughts

• $\frac{\langle (1,0,0)^T,(0,1,0)^T,(0,0,1)^T\rangle}{\langle (1,4,7)^T,(2,5,8)^T\rangle }$ is the same as $\frac{\langle (1,4,7)^T,(0,1,0)^T,(0,0,1)^T\rangle}{\langle (1,4,7)^T,(2,5,8)^T\rangle }$ (As [ilmath](1,0,0)[/ilmath] is in the span of this, and [ilmath](1,4,7)[/ilmath] is clearly in the span of the standard basis vectors)