hcju/proofwikiqa huggingface dataset:
Let $\triangle ABC, \triangle DCE$ be two triangles such that $BA : AC = DC : DE$ be situated so that $AB \parallel DC$ and $AC \parallel DE$. We need to show that $BC$ is in a straight line with $CE$. :300px We have that $AB \parallel DC$. So from Parallelism implies Equal Alternate Angles we have that $\angle BAC = \angle ACD$. For the same reason $\angle CDE = \angle ACD$. So $\angle BAC = \angle CDE$. From Triangles with One Equal Angle and Two Sides Proportional are Similar, it follows that $\triangle ABC$ is similar to $\triangle DCE$. Therefore $\angle ABC = \angle DCE$. It follows that $\angle ACE = \angle ABC + \angle BAC$. Add $\angle ACB$ to each. Then $\angle ACE + \angle ACB = \angle BAC + \angle ACB + \angle CBA$. But from Sum of Angles of Triangle Equals Two Right Angles $\angle BAC + \angle ACB + \angle CBA$ equals two right angles. So $\angle ACE + \angle ACB$ equals two right angles. The result follows from Two Angles making Two Right Angles make Straight Line. {{qed}} {{Euclid Note|32|VI}}
Let $ABC, DEF$ be equal circles. Let $\angle BGC, \angle EHF$ be angles at their centres $G, H$. Let $\angle BAC, \angle EDF$ be angles at their circumferences. We need to show that: : $BC : EF = \angle BGC : \angle EHF = \angle BAC : \angle EDF$ :400px Let any number of consecutive arcs $CK, KL$ be made equal to the arc $BC$. Also let any number of consecutive arcs $FM, MN$ be made equal to the arc $EF$. Join $GK, GL, HM, HN$. Then since $BC = CK = KL$ it follows from Angles on Equal Arcs are Equal that: : $\angle BGC = \angle CGK = \angle KGL$ Therefore, whatever multiple the arc $BL$ is of $BC$, that multiple is also the angle $\angle BGL$ of the angle $\angle BGC$. For the same reason, whatever multiple the arc $NE$ is of $EF$, that multiple is also the angle $\angle NHE$ of the angle $\angle EHF$. Then from Angles on Equal Arcs are Equal: : $BL = EN \implies \angle BGL = \angle EHN$ : $BL > EN \implies \angle BGL > \angle EHN$ : $BL < EN \implies \angle BGL < \angle EHN$ Then from {{EuclidDefLink|V|5|Equality of Ratios}}: : $BC : EF = \angle BGL : \angle EHF$ But from the Inscribed Angle Theorem: : $\angle BGC : \angle EHF = \angle BAC : \angle DEF$ Hence the result. {{qed}} {{Euclid Note|33|VI}}
Let the less of two unequal numbers $AB, CD$ be continually subtracted from the greater, such that the number which is left over never measure the one before it till a unit is left. We need to show that $AB$ and $CD$ are coprime. :200px Suppose $AB, CD$ are not coprime. Then some natural number $E > 1$ will divide them both. Let some multiple of $CD$ be subtracted from $AB$ such that the remainder $AF$ is less than $CD$. Then let some multiple of $AF$ be subtracted from $CD$ such that the remainder $CG$ is less than $AF$. Then let some multiple of $CG$ be subtracted from $FA$ such that the remainder $AH$ is a unit. Since, then, $E$ divides $CD$, and $CD$ divides $BF$, then $E$ also divides $BF$. But $E$ also divides $BA$. Therefore $E$ also divides $AF$. But $AF$ divides $DG$. Therefore $E$ also divides $DG$. But $E$ also divides the whole $DC$. Therefore $E$ also divides the remainder $GC$. But $CG$ divides $FH$. Therefore $E$ also divides $FH$. But $E$ also divides the whole $FA$. Therefore $E$ also divides the remainder, that is, the unit $AH$. But $E > 1$ so this is impossible. Therefore, from {{EuclidDefLink|VII|12|Relatively Prime}}, $AB$ and $CD$ are relatively prime. {{qed}} {{Euclid Note|1|VII}}
If $\alpha = 0$ then any integer works and the proof is finished. Let $\alpha \ne 0$. Let $\map f x = x^d + a_{d - 1} x^{d - 1} + \dotsb + a_0$ be the minimal polynomial of $\alpha$ over $\Q$.  Suppose that $a_i = \dfrac {b_i} {c_i}$ is a reduced fraction for each $i$ such that $a_i \ne 0$. Let $n$ be the least common multiple of the $c_i$, of which there must be at least one by our assumptions.  Consider the polynomial: {{begin-eqn}} {{eqn | l = \map g x       | r = n^d \map f {\frac x n}       | c =  }} {{eqn | r = n^d \paren {\frac {x^d} {n^d} + a_{d - 1} \frac {x^{d - 1} } {n^{d - 1} } + \dotsb + a_0}       | c =  }} {{eqn | r = x^d + n a_{d - 1} x^{d - 1} + \dotsb + n^d a_0       | c =  }} {{end-eqn}} Note that $g$ is a monic polynomial with coefficients in $\Z$ by our choice of $n$. Furthermore, by construction, we see that $n \alpha$ is a root of $g$ and is therefore an algebraic integer. {{qed}} Category:Number Theory Category:Algebraic Number Theory t6uf3u1n7iqu3slikfj66ht5frq2kd3
Let $AB, CD$ be the two given (natural) numbers which are not coprime. We need to find the greatest common divisor of $AB$ and $CD$. :250px {{WLOG}} that $CD \le AB$. We have that $CD$ is a divisor of itself. If $CD$ is a divisor of $AB$ then $CD$ is a common divisor of $CD$ and $AB$. It is clearly the greatest, because no number greater than $CD$ can be a divisor  of $CD$. Now, suppose $CD$ does not divide $AB$. Then the less of the numbers $AB, CD$ being continually subtracted from the greater, some number will be left which will be a divisor  of the one before it. From Coprimality Criterion, this number will not be a unit, otherwise $AB$ and $CD$ will be coprime. So some number will be left which is a divisor of the one before it. Now let $CD$, dividing $BE$, leave $EA$ less than itself. Let $EA$, dividing $DF$, leave $FC$ less than itself. Let $CF$ divide $AE$. Since then $CF$ divides $AE$, and $AE$ divides $DF$, then $CF$ will also divide $DF$. But it also divides itself. Therefore it will also divide the whole $CD$. But $CD$ divides $BE$, therefore $CF$ divides $BE$. But it also divides $EA$, therefore it will also divide the whole $BA$. So $CF$ is a common divisor of $CD$ and $AB$. Suppose $CF$ is not the greatest common divisor of $CD$ and $AB$. Let $G > CF$ also be a common divisor of $CD$ and $AB$. Since $G$ divides $CD$, while $CD$ divides $BE$, it follows that $G$ divides $BE$. But $G$ also divides the whole $BA$, and so it also divides the remainder $AE$. But $AE$ divides $DF$. Therefore $G$ divides $DF$. But $G$ also divides the whole $DC$. Therefore it will also divide the remainder $CF$. But $G$ is greater than $CF$, which is impossible. Therefore no number greater than $CF$ divides the numbers $AB$ and $CD$. Therefore $CF$ is the greatest common divisor of $AB$ and $CD$. {{qed}}
By Gauss's Lemma on Irreducible Polynomials, it suffices to show that $f$ is irreducible in $\Z \sqbrk x$. {{AimForCont}} that $f = g h$ where $g, h \in \Z \sqbrk x$ are both non-constant. Let: :$\map g x = b_e x^e + b_{e - 1} x^{e - 1} + \dotsb + b_0$ :$\map h x = c_f x^f + c_{f - 1} x^{f - 1} + \dotsb + c_0$ Then we have for each $i$: :$\displaystyle a_i = \sum_{j + k \mathop = i} {b_j c_k}$ In particular, it follows that: :$a_0 = b_0 c_0$ Possibly after exchanging $g$ and $h$, we may arrange that: :$p \nmid c_0$ by condition $(2)$. From condition $(1)$, it follows that then necessarily: :$p \divides b_0$ We also have: :$a_d = b_e c_f$ and by condition $(1)$: :$p \nmid a_d$ and hence: :$p \nmid b_e$ It follows that there exists a smallest positive $i$ such that: :$p \nmid b_i$ Naturally, $i \le e$.  By assumption, both $g$ and $h$ are non-constant. Hence by Degree of Product of Polynomials over Integral Domain: :$i < d$ {{explain|Re-evaluate the above link - may need to be Degree of Product of Polynomials over Integral Domain not Less than Degree of Factors. Clarification needed.}} Consider: :$a_i = b_0 c_i + b_1 c_{i - 1} + \dotsb + b_i c_0$ with the convention that $c_j = 0$ if $j > f$. By the minimality of $i$, it follows that: :$p \divides b_k$ for $0 \le k < i$. Also, since neither $c_0$ nor $b_i$ is divisible by $p$, the last term $b_i c_0$ is '''not''' divisible by $p$. Thus, we conclude that: :$p \nmid a_i$ which contradicts condition $(1)$. Therefore, $f$ is irreducible. {{qed}}
Let $A, B, C$ be the three given (natural) numbers not prime to one another. :250px Let the GCD $D$ of $A, B$ be found, by the Euclidean Algorithm. Then $D$ either divides or does not divide $C$. First, suppose $D$ divides $C$. Then as it also divides $A$ and $B$, it is a common divisor of $A, B$ and $C$. Suppose $D$ is not the GCD of $A, B, C$. Then some number $E$ is a common divisor of $A, B$ and $C$ such that $E > D$. But then $E$ is a common divisor of $A$ and $B$ and so from the porism to Euclid's Algorithm $E$ also divides $D$. That means $E \le D$ which contradicts $E > D$. So $D$ is the GCD of $A, B, C$ after all. Next, suppose $D$ is not a divisor of $C$. Since $A, B, C$ are not prime to one another, some number will divide them. That number which divides $A, B, C$ will also divide $A, B$ and so from the porism to Euclid's Algorithm will also divide $D$. But it also divides $C$. So some number divides both $D$ and $C$, and so $C$ and $D$ are not prime to one another. Let the GCD of $C$ and $D$ be $E$, by the Euclidean Algorithm. Since $E$ divides $D$ and $D$ divides $A, B$, then $E$ divides $A, B$. But $E$ divides $C$, and so $E$ divides $A, B, C$. So $E$ is a common divisor of $A, B$ and $C$. Now suppose $E$ is not the GCD of $A, B, C$. Then some number $F$ such that $F > E$ divides $A, B, C$. It follows that $F$ divides $A$ and $B$, and so from the porism to Euclid's Algorithm will also divide $D$, the GCD of $A, B$. So $F$ divides $D$ and $F$ also divides $C$. So $F$ is a common divisor of $D$ and $C$. So by the porism to Euclid's Algorithm, $F$ divides the GCD of $C, D$, which is $E$. But this contradicts the supposition that $F > E$. So no such $F$ can exist, and so $E$ is the GCD of $A, B, C$. {{Qed}} {{Euclid Note|3|VII}}
Let $A, BC$ be two (natural) numbers and let $BC < A$. We need to show that $BC$ is either an aliquot part or aliquant part of $A$. That is, either $BC$ is a divisor of $A$, or it is a multiple of some divisor of $A$. :200px $A$ and $BC$ are either coprime or they are not. First, suppose $A$ and $BC$ are coprime. Then if $BC$ be divided into the units in it, each unit of $BC$ will be some aliquot part of $A$, so that $BC$ is an aliquant part of $A$. Next, let $A$ and $BC$ not be coprime. Then $BC$ either divides $A$ or it does not. If $BC$ divides $A$ then $BC$ is an aliquot part of $A$. But if not, then let $D$ be the GCD of $A$ and $BC$ by Euclid's algorithm. Let $BC$ be divided into the numbers equal to $D$, namely $BE, EF, FC$. Then each of the numbers $BE, EF, FC$ is also an aliquot part of $A$. That is, $BC$ is an aliquant part of $A$. {{qed}} {{Euclid Note|4|VII}}
Let $x \in \map  \cl H$ be an isolated point of $\map \cl H$. {{AimForCont}} that $x$ is not an isolated point of $H$. Then by the definition of a limit point, it follows that $x$ must be a limit point of $H$. From Set is Subset of its Topological Closure: :$H \subseteq \map \cl H$ So by Limit Point of Subset is Limit Point of Set it follows that $x$ is a limit point of $\map \cl H$. But then $x$ cannot be an isolated point of $\map \cl H$. This is a contradiction. Therefore every isolated point of $\map \cl H$ is also an isolated point of $H$. {{qed}}
Let $x \in \map \cl H$ be isolated in $\map \cl H$. By definition of isolated point, $x$ is not a limit point of $\map \cl H$. From Set is Subset of its Topological Closure: :$H \subseteq \map \cl H$ We have that Limit Point of Subset is Limit Point of Set. But $x$ is not a limit point of $\map \cl H$ . So by the Rule of Transposition, $x$ cannot be a limit point of $H$. As $x$ is not a limit point of $H$, it follows that $x$ must be an isolated point of $H$. {{qed}}
Let the (natural) number $A$ be an aliquot part of the (natural) number $BC$. Let the (natural) number $D$ be the same aliquot part of another (natural) number $EF$ that $A$ is of $BC$. We need to show that $A + D$ is the same aliquot part of $BC + EF$. :250px We have that whatever aliquot part $A$ is of $BC$, then $D$ is also the same aliquot part of $EF$. Therefore as many numbers as there are in $BC$ equal to $A$, so many numbers are there in $EF$ equal to $D$. Let $BC$ be divided into the numbers equal to $A$, that is, $BG, GC$. Let $EF$ be divided into the numbers equal to $D$, that is, $EH, HF$. Then the multitude of $BG, GC$ will be equal to the multitude of $EH, HF$. Since $BG = A$ and $EH = D$, then $BG + EH = A + D$. For the same reason, $GC + HF = A + D$. Therefore, as many numbers as there are in $BC$ equal to $A$, so many are there also in $BC + EF$ equal to $A + D$. Therefore, whatever multiple $BC$ is of $A$, the same multiple also is $BC + EF$ of $A + D$. Therefore, whatever aliquot part $A$ is of $BC$, the same aliquot part also is $A + D$ of $BC + EF$. {{qed}}
A direct application of the Distributive Property: :$\dfrac 1 n b + \dfrac 1 n d = \dfrac 1 n \paren {b + d}$ {{qed}}
A direct application of the Distributive Property: :$\dfrac m n b + \dfrac m n d = \dfrac m n \paren {b + d}$ {{qed}}
Let the (natural) number $AB$ be an aliquant part of the (natural) number $C$. Let the (natural) number $DE$ be the same aliquant part of another (natural) number $F$ that $AB$ is of $C$. We need to show that $AB + DE$ is the same aliquant part of $C + F$. :250px We have that whatever aliquant part $AB$ is of $C$, $DE$ is also the same aliquant part as $F$. It follows that as many aliquant parts of $C$ as there are in $AB$, so many aliquant parts of $F$ are there also in $DE$. Let $AB$ be divided into the aliquant parts of $C$, namely $AG, GB$, and $DE$ into the aliquant parts of $F$, namely $DH, HE$. Thus the multitude of $AG, GB$ will be equal to the multitude of $DH, HE$. We have that whatever aliquot part $AG$ is of $C$, the same aliquot part is $DH$ of $F$ also. Therefore, from {{EuclidPropLink|book = VII|prop = 5|title = Divisors obey Distributive Law}}, whatever aliquot part $AG$ is of $C$, the same aliquot part also is $AG + DH$ of $C + F$ also. For the same reason, whatever aliquot part $GB$ is of $C$, the same aliquot part also is $GB + HE$ of $C + F$. Therefore whatever aliquant part $AB$ is of $C$, the same aliquant part also is $AB + DE$ of the $C + F$ {{qed}}
Since $B \in \tau$, it follows by definition that $S \setminus B$ is closed. Thus: {{begin-eqn}} {{eqn | l = A \cap B       | r = \varnothing }} {{eqn | ll= \implies       | l = A       | o = \subseteq       | r = S \setminus B       | c = Intersection with Complement is Empty iff Subset }} {{eqn | ll= \implies       | l = A^-       | o = \subseteq       | r = \left({S \setminus B}\right)^-       | c = Topological Closure of Subset is Subset of Topological Closure }} {{eqn | ll= \implies       | l = A^-       | o = \subseteq       | r = S \setminus B       | c = Closed Set Equals its Closure, and $S \setminus B$ is closed }} {{eqn | ll= \implies       | l = A^- \cap B       | r = \varnothing       | c = Intersection with Complement is Empty iff Subset }} {{end-eqn}} {{qed}} Category:Set Closures Category:Open Sets oqcembtzowkq0muwxw3tcjsla77fvmn
Let $f$ and $g$ be as follows: {{begin-eqn}} {{eqn | l = f       | r = \sum_{k \mathop \in \Z} a_k X^k }} {{eqn | l = g       | r = \sum_{k \mathop \in \Z} b_k X^k }} {{end-eqn}} From the definition of primitive polynomial, the coefficients of $f$ and $g$ are all integers. From the definition of polynomial product: :$\displaystyle f g = \sum_{k \mathop \in \Z} c_k \mathbf X^k$  where: :$\displaystyle c_k = \sum_{\substack {p \mathop + q \mathop = k \\ p, q \mathop \in \Z}} a_p b_q$ it is clear that the coefficients of $f g$ are also all integers. {{AimForCont}} $f g$ is not primitive. Then its coefficients must have a GCD greater than $1$. Therefore there exists some prime $p$ which divides all the coefficients of $fg$. Now $p$ can not divides all the coefficients of either $f$ or $g$, because they are primitive polynomials. So: :Let $i$ be the smallest integer such that $p$ does not divide $a_i$ :Let $j$ be the smallest integer such that $p$ does not divide $b_j$. Consider the coefficient $c_{i+j}$ of $f g$: :$\displaystyle c_{i + j} = \sum_{k \mathop = 0}^{i + j} a_k b_{i + j - k} = a_0 b_{i + j} + a_1 b_{i + j - 1} + \cdots + a_i b_j + \cdots + a_{i + j - 1} b_1 + a_{i + j} b_0$ From the assumption, it follows that $p \divides c_{i + j}$, and so: :$\displaystyle p \divides \sum_{k \mathop = 0}^{i + j} a_k b_{i + j - k}$ where $\divides$ denotes divisibility. Also, from the choice of $i$ and $j$, we have: :$p \divides a_m$ whenever $m < i$ :$p \divides b_n$ whenever $n < j$. Now all the terms of $\displaystyle \sum_{k \mathop = 0}^{i + j} a_k b_{i + j - k}$, except for $a_i b_j$, contain a factor from either $\set {a_0, a_1, \ldots, a_{i - 1} }$ or $\set {b_0, b_1, \ldots, b_{j - 1} }$. It follows that we have: :$p \divides \paren {\displaystyle \sum_{k \mathop = 0}^{i - 1} a_k b_{i + j - k} + \sum_{k \mathop = i + 1}^{i + j} a_k b_{i + j - k} }$ But then, as also $p \divides c_{i + j}$, it follows that $p \divides a_i b_j$ as well. From Euclid's Lemma for Prime Divisors, $p$ has to divide one or the other. This contradicts the definition of $i$ and $j$. So $i$ and $j$ cannot both exist. It follows that $p$ divides at least one of $f$ and $g$, one of which is therefore not primitive. From this contradiction, we conclude that $f g$ must be primitive. Hence the result. {{qed}}
Recall Polynomial has Integer Coefficients iff Content is Integer. By hypothesis $f$ and $g$ have content $1 \in \Z$. Therefore: :$f, g \in \Z \sqbrk X$ {{AimForCont}} that $f g$ is not primitive, say: :$\cont {f g} = d \ne 1$ By the Fundamental Theorem of Arithmetic we can choose a prime $p$ dividing $d$. Let $\pi : \Z \to \Z / p \Z$ be the canonical epimorphism to the ring of integers modulo $p$. From Ring of Integers Modulo Prime is Field, $\Z / p \Z$ is a field. Let $\Pi : \Z \sqbrk X \to \paren {\Z / p \Z} \sqbrk X$ be the induced homomorphism of the polynomial rings. By construction, $p$ divides each coefficient of $f g$, so: :$\map \Pi {f g} = \map \Pi f \, \map \Pi g = 0$ From Polynomial Forms over Field form Integral Domain, $\paren {\Z / p \Z} \sqbrk X$ is an integral domain. Thus: :$\map \Pi f = 0$ or $\map \Pi g = 0$ After possibly exchanging $f$ and $g$, we may assume that $\map \Pi f = 0$. Now by Kernel of Induced Homomorphism of Polynomial Forms, if: :$f = a_0 + a_1 X + \cdots + a_n X^n$ we must have: :$\map \pi {a_i} = 0$ for $i = 0, \ldots, n$. That is: :$a_i \in p \Z$ for $i = 0, \ldots, n$. But this says precisely that $p$ divides each $a_i$, $i = 0, \ldots, n$. Therefore $p$ divides the content of $f$, a contradiction. Hence our assumption that $f g$ is not primitive was invalid. The result follows by Proof by Contradiction. {{Qed}}
First note that Modulo Addition is Well-Defined. Taking the group axioms in turn: === G0: Closure === In Real Number minus Floor it is demonstrated that: :$\forall x, y \in S: x \circ y \in S$ Thus $\left({S, \circ}\right)$ is closed. {{qed|lemma}} === G1: Associativity === The associativity of $\circ$ follows from that of the sum of real numbers. {{finish|More needs to be done here.}} Thus $\circ$ is associative on $S$. {{qed|lemma}} === G2: Identity === By definition of $S$: :$0 \in S$ Let $x \in S$. We have that: :$0 \le x < 1$ and so by definition of floor function: :$\left\lfloor{x}\right\rfloor = 0$ So: {{begin-eqn}} {{eqn | l = x \circ 0       | r = x + 0 - \left\lfloor{x + 0}\right\rfloor       | c =  }} {{eqn | r = x - \left \lfloor{x}\right\rfloor       | c =  }} {{eqn | r = x - 0       | c = from above }} {{eqn | r = x       | c =  }} {{end-eqn}} Hence $\left({S, \circ}\right)$ has $0$ as an identity element. {{qed|lemma}} === G3: Inverses === Let $x \in S$. First let $x = 0$. We have that $0$ is the identity of $\left({S, \circ}\right)$: :$0 \circ 0 = 0$ and so from Inverse of Identity Element is Itself, $0$ is its own inverse. Now let $x \ne 0$. By definition of $S$: : $0 \le x < 1$ Hence: : $1 - x \in S$ (Note that because $1 - 0 \notin S$, the above is not true for $x = 0$, which is why it has been treated as a special case.) Therefore: {{begin-eqn}} {{eqn | l = x \circ \left({1 - x}\right)       | r = x + 1 - x - \left\lfloor{x + 1 - x}\right\rfloor       | c =  }} {{eqn | r = 1 - \left\lfloor{1}\right\rfloor       | c =  }} {{eqn | r = 1 - 1       | c = Real Number is Integer iff equals Floor }} {{eqn | r = 0       | c =  }} {{end-eqn}} From G3: Identity, $0$ is the identity of $\left({S, \circ}\right)$. Thus every element $x$ of $\left({S, \circ}\right)$ has an inverse: : $\begin{cases} 1 - x & : x \ne 0 \\ 0 & : x = 0 \end{cases}$ {{qed|lemma}} All the group axioms are thus seen to be fulfilled, and so $\left({S, \circ}\right)$ is a group. {{qed}} Category:Examples of Groups fsmvdwzrcl9bz2zqlbe7f4vhuswt2rc
Let $x \in S$. Then $\phi \left({x}\right) \in T$. Since $\mathcal U$ is a cover of $T$: :$\exists U \in \mathcal U: \phi \left({x}\right) \in U$ By definition of preimage, $x \in \phi^{-1} \left({U}\right)$. So: :$\forall x \in S: \exists \phi^{-1} \left({U}\right) \in S: x \in \phi^{-1} \left({U}\right)$ That is, $\left\{{\phi^{-1} \left({U}\right): U \in \mathcal U}\right\}$ is a cover of $S$. {{qed}} Category:Preimages under Mappings Category:Covers s9i6e7rhfl75r715x5gsczgjl4cbz86
Let $T_A$ be compact. Take an open cover $\UU$ of $T_B$. From Preimage of Cover is Cover, $\set {\phi^{-1} \sqbrk U: U \in \UU}$ is a cover of $S_A$. But $\phi$ is continuous, and for all $U \in \UU$, $U$ is open in $T_B$. It follows that $\forall U \in \UU: \phi^{-1} \sqbrk U$ is open in $T_A$. So $\set {\phi^{-1} \sqbrk U:\ U \in \UU}$ is an open cover of $T_A$. $T_A$ is compact, so we take a finite subcover: :$\set {\phi^{-1} \sqbrk {U_1}, \ldots, \phi^{-1} \sqbrk {U_n} }$ We have that $\phi$ is surjective. So from Surjection iff Right Inverse, $\phi \sqbrk {\phi^{-1} \sqbrk A} = A$ So: :$\set {\phi \sqbrk {\phi^{-1} \sqbrk {U_1} }, \ldots, \phi \sqbrk {\phi^{-1} \sqbrk {U_n} } } = \set {U_1, \ldots, U_n} \subseteq \UU$ is a finite subcover of $\UU$ on $T_B$. {{qed}}
Let $T_A$ be $\sigma$-compact. Then: :$\displaystyle S_A = \bigcup_{i \mathop = 1}^\infty S_i$ where $S_i \subseteq S_A$ are compact. Since $\phi$ is surjective, we have from Image of Union under Relation: :$\displaystyle \phi \sqbrk {S_A} = S_B = \phi \sqbrk {\bigcup_{i \mathop = 1}^\infty S_i} = \bigcup_{i \mathop = 1}^\infty \phi \sqbrk {S_i}$ From Compactness is Preserved under Continuous Surjection, we have that $\phi \sqbrk {S_i}$ is compact for all $i \in \N$. So $S_B$ is the union of a countable number of compact subsets. Thus, by definition, $T_B$ is also $\sigma$-compact. {{qed}}
Let $T_A$ be countably compact. Take a countable open cover $\UU$ of $T_B$. From Preimage of Cover is Cover, $\VV := \set {\phi^{-1} \sqbrk U: U \in \UU}$ is a cover of $S_A$. $\VV$ is a countable cover because there it is bijective with $\UU$. {{explain|It needs to be demonstrated that "it" is bijective, and in this context, exactly what "it" is.}} By hypothesis, $\phi$ is continuous. For all $U \in \UU$, $U$ is open in $T_B$. It follows that $\forall U \in \UU: \phi^{-1} \sqbrk U$ is open in $T_A$. So $\VV$ is a countable open cover of $T_A$. $T_A$ is countably compact, so we take a finite subcover: :$\set {\phi^{-1} \sqbrk {U_1}, \ldots, \phi^{-1} \sqbrk {U_n} }$ We have that $\phi$ is surjective. So from Surjection iff Right Inverse: :$\phi \sqbrk {\phi^{-1} \sqbrk A} = A$ So: :$\set {\phi \sqbrk {\phi^{-1} \sqbrk {U_1} }, \ldots, \phi \sqbrk {\phi^{-1} \sqbrk {U_n} } } = \set {U_1, \ldots, U_n} \subseteq \UU$ is a finite subcover of $\UU$ on $T_B$. {{explain|It is not clear that $\set {U_1, \ldots, U_n}$ covers $T_B$.}} {{qed}}
Let $T_A$ be a sequentially compact space. Take an infinite sequence $\sequence {x_n} \subseteq S_B$. From the surjectivity of $\phi$, there exists another infinite sequence $\sequence {y_n} \subseteq S_A$ such that $\map \phi {y_n} = x_n$. By the definition of sequential compactness, there exists a subsequence of $\sequence {y_n}$ Let this subsequence be named $\sequence {y_{n_k} }$. Let $\sequence {y_{n_k} }$ converge to $y \in S_A$ with respect to $T_A$. From the continuity of $\phi$, it is concluded that $\map \phi {y_{n_k} } = x_{n_k}$ converges to $\map \phi y \in S_B$. Thus, $\sequence {x_n}$ has a subsequence that converges. By definition, $T_B$ is sequentially compact. {{qed}}