# Matt N.

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 13h comment $f$ is one-to-one with domain $\mathbb{R} - \{ a\}$ implies range $\mathbb{R} - \{ b\}$? @Unwisdom Thank you for your comment. I suppose the problem arises when, after mapping $(-\infty,a)$ to $(-\infty,a)$, one is forced to map $(a,\infty)$ to $[a,\infty)$. 23h comment $f$ is one-to-one with domain $\mathbb{R} - \{ a\}$ implies range $\mathbb{R} - \{ b\}$? @Unwisdom Thank you. I had a long day, could you please elaborate a bit? I know that continuous maps map connected sets to connected sets but it's not obvious to me this very second why this implies that they also have to map disconnected sets to disconnected sets. 23h comment $f$ is one-to-one with domain $\mathbb{R} - \{ a\}$ implies range $\mathbb{R} - \{ b\}$? Is it also possible if $f$ is required to be continuous in addition to being bijective? Mar4 awarded Nice Answer Mar1 comment What is the order of $(\mathbb{Z} \oplus \mathbb{Z})/ \langle (2,2) \rangle$ and is it cyclic? Ooh, you are right, I missed lots of copies of $\mathbb Z$. Thank you very much for your patient comments. I do not fully understand your comment somehow but it gave me the idea on how to prove that the image is $\mathbb Z \oplus \mathbb Z_2$: define a map $\mathbb Z\oplus\mathbb Z \to \mathbb Z\oplus\mathbb Z$ that maps $\langle (0,2) \rangle$ to $(0,0)$ and whose image is $\mathbb Z \oplus \mathbb Z_2$. Like e.g. $f((n,k)) = (n, k \mod 2)$. My only problem now is that I don't know how to figure out what the image looks like. How did you figure out it's $\mathbb Z \oplus \mathbb Z_2$? Feb28 comment What is the order of $(\mathbb{Z} \oplus \mathbb{Z})/ \langle (2,2) \rangle$ and is it cyclic? Just a quick question to see if I am "seeing" the groups in this question correctly: The group $\mathbb Z \oplus \mathbb Z$ contains $3$ copies of $\mathbb Z$ -- one looks like $(k,0)$, one like $(0,k)$ and one like $(k,k)$. Then taking the quotient by $\langle (2,2) \rangle$ eliminates half of the copy $(k,k)$ (namely the even numbers). So what remains is all elements of $\mathbb Z \oplus \mathbb Z$ minus all elements of the form $(2n,2n)$? Feb24 comment Counting roots of polynomial inside $S^1$ @DanielFischer Yes, thanks. I was 99% sure it wasn't but... you (=I) never know. : ) I think I understand: It looks like if $a_n$ are the zeros of $p$ then the fraction ${p' \over p}$ expands as a sum $\sum_n {k_n \over z-a}$. Computing the integral of the sum yields $\sum_n k_n$ because $\oint {1 \over z-a}dz$ around $a$ is $1$. It's pretty neat. Feb24 comment Counting roots of polynomial inside $S^1$ @DanielFischer That $q$ in your first comment is not the same $q$ as in the question, right? Feb22 comment Concatenation with continuous function is entire Thank you for your comment. I'm still a bit confused. I thought $\varphi$ was the linear operator $A_x$ here and $A_x$ is the derivative of $f$ at $x$. But then we'd have $\varphi = A_x = f'$ at $x$.Which of the equalities here I thought were true does not hold? Feb22 comment Concatenation with continuous function is entire Did you mean $f'(a) =\varphi$ for all $a \in A$? (in the last sentence before the third quotation) Or perhaps $f'(a) =\varphi(a)$? Feb21 comment Continuity of the derivative @Etienne Thank you for your comment. Indeed a seemingly not very thought out comment of mine, I'm sorry about that. Especially if one thinks about real functions where the linear map that is the derivative becomes the tangent! Feb21 comment Continuity of the derivative @Etienne I am not sure if I understand your question correctly. If one considers $f: \mathbb C \to \mathbb C$ as a function from $f: \mathbb R^2 \to \mathbb R^2$ then if $f$ is (complex) differentiable at a point $x$ its derivative at $x$ is given by the Jacobian of $f$. But in finite dimension every linear map is continuous which would yield what I understand you are asking. But I think I misunderstand your question or I am probably making a fundamental mistake in what I wrote. Feb21 revised Continuity of the derivative Some very minor spelling corrected. Feb20 revised Counting roots of polynomial inside $S^1$ edited tags; edited title Feb20 answered Counting roots of polynomial inside $S^1$ Feb19 revised A closed ideal in a commutative Banach algebra $C(X)$ FA tag added. Feb19 comment A closed ideal in a commutative Banach algebra $C(X)$ @DanielFischer I went ahead and made your comments into an answer. Hope it's ok. Feb19 answered A closed ideal in a commutative Banach algebra $C(X)$ Feb18 revised Vector fields on $\Bbb S^2$ added 1 characters in body Feb18 comment Cycling Digits puzzle @DavidConrad Yes the number is assumed to be in base $10$.