hcju/proofwikiqa huggingface dataset:
We have Identity Mapping is Bijection. We also have Identity Mapping is Continuous. Hence, by definition, $I_T$ is a homeomorphism. {{qed}} Category:Homeomorphisms Category:Identity Mappings 5s2tqa0d7guhqxk08qpyx4u63jif5wa
By definition of homeomorphism, $f$ and $g$ are both bijections. From Composite of Bijections is Bijection it follows that $g \circ f$ is also a bijection. Similarly, also by definition of homeomorphism, $f$ and $g$ are both continuous mappings. From Composite of Continuous Mappings is Continuous it follows that $g \circ f$ is also a continuous mapping. Hence the result, from definition of homeomorphism. {{qed}} Category:Homeomorphisms 6wti8nffp909u0kceehmavp8m0shjq1
Let $U \in \tau$. We have Identity Mapping is Bijection. So $I_S^{-1}$ is well-defined and: :$\forall x \in U: \map {I_S^{-1} } x = x$ Thus $I_S^{-1} \sqbrk U = U \in \tau$. Hence, by definition of continuous mapping, $I_S$ is continuous. {{qed}}
Let $U \in \tau$ be open in $T$. Then by definition of finite complement topology, $S \setminus U$ is finite. By definition of open set, $S \setminus U$ is closed. As $S$ is infinite, it follows that $U$ must also be infinite. Thus unless $U = S$, $S \setminus U$ can not be open. Hence the result. {{qed}}
We have by definition that $\varnothing \in \tau_H$, and as $H \subseteq S$ we have that $S \in \tau_H$. Now let $U_1, U_2 \in \tau_H$. By definition $H \subseteq U_1 $ and $H \subseteq U_2$. From Intersection is Largest Subset it follows that $H \subseteq \left({U_1 \cap U_2}\right)$. So $U_1 \cap U_2 \in \tau_H$. Now let $\mathcal U \subseteq \tau_H$. We have by definition that $\forall U \in \mathcal U: H \subseteq U$. Hence from Subset of Union $H \subseteq \bigcup \mathcal U$. So all the properties are fulfilled for $\tau_H$ to be a topology on $S$. {{qed}} Category:Included Set Topology tcccaqqmlgqp9o29glyh206padtt3m3
Let $T_1 = \struct {S, \tau_{A_1} }$ and $T_2 = \struct {S, \tau_{A_2} }$ be included set spaces on $S$. === Necessary Condition === Let $T_1 \ge T_2$. We have that $A_2 \in \tau_{A_2}$ by definition of included set topology. But by definition of finer topology, $A_2 \in \tau_{A_1}$. So by definition of included set topology, $A_1 \subseteq A_2$. Now suppose that $T_1 > T_2$. That is, $T_1 \ge T_2$ and $T_1 \ne T_2$. It follows that $\exists X \in \tau_{A_1}: X \notin \tau_{A_2}$. Then  $A_1 \subseteq X \subsetneq A_2$ from which it follows that $A_1 \subsetneq A_2$. {{qed|lemma}} === Sufficient Condition === Let $A_1 \subseteq A_2$. Then: {{begin-eqn}} {{eqn | l = U       | o = \in       | r = \tau_{A_2}       | c =  }} {{eqn | ll= \leadsto       | l = A_2       | o = \subseteq       | r = U       | c = {{Defof|Included Set Topology}} }} {{eqn | ll= \leadsto       | l = A_1       | o = \subseteq       | r = U       | c = Subset Relation is Transitive }} {{eqn | ll= \leadsto       | l = U       | o = \in       | r = \tau_{A_1}       | c = {{Defof|Included Set Topology}} }} {{end-eqn}} So $T_1 \ge T_2$ by definition of finer topology. Now suppose that $A_1 \subsetneq A_2$. Then $A_2 \not \subseteq A_1$ and so $A_1 \notin \tau_{A_2}$. Thus $\tau_{A_1} \not \subseteq \tau_{A_2}$ and so $T_1 > T_2$. {{qed|lemma}} Both necessity and sufficiency have been proved, hence the result. {{qed}}
We are given that $\mathbf A$ and $\mathbf B$ are invertible. From Product of Matrices is Invertible iff Matrices are Invertible, $\mathbf A \mathbf B$ is also invertible. By the definition of inverse matrix: :$\mathbf A \mathbf A^{-1} = \mathbf A^{-1} \mathbf A = \mathbf I$ and  :$\mathbf B \mathbf B^{-1} = \mathbf B^{-1} \mathbf B = \mathbf I$ Now, observe that: {{begin-eqn}} {{eqn | l = \paren {\mathbf A \mathbf B} \paren {\mathbf B^{-1} \mathbf A^{-1} }       | r = \paren {\mathbf A \paren {\mathbf B \mathbf B^{-1} } } \mathbf A^{-1}       | c = Matrix Multiplication is Associative }} {{eqn | r = \paren {\mathbf A \mathbf I} \mathbf A^{-1} }} {{eqn | r = \mathbf A \mathbf A^{-1}       | c = {{Defof|Identity Element}} }} {{eqn | r = \mathbf I }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = \paren {\mathbf B^{-1} \mathbf A^{-1} } \paren {\mathbf A \mathbf B}       | r = \paren {\mathbf B^{-1} \paren {\mathbf A^{-1} \mathbf A} } \mathbf B       | c = Matrix Multiplication is Associative }} {{eqn | r = \paren {\mathbf B^{-1} \mathbf I} \mathbf B }} {{eqn | r = \mathbf B^{-1} \mathbf B       | c = {{Defof|Identity Element}} }} {{eqn | r = \mathbf I }} {{end-eqn}} The result follows from the definition of inverse. {{qed}}
Let $s, t \in \closedint 1 m$ such that $s \ne t$. === Case $1$ === Let $e$ be the elementary row operation $r_s \to \lambda r_s$: :$E_{ik} = \begin{cases} \delta_{ik} & : i \ne s \\ \lambda \delta_{ik} & : i = s \end{cases}$ where $\delta$ denotes the Kronecker delta. Then: {{begin-eqn}} {{eqn | l = \sqbrk {E A}_{i j}       | r = \sum_{k \mathop = 1}^m E_{i k} A_{k j}       | c =  }} {{eqn | r = \begin{cases} A_{i j} & : i \ne r \\ \lambda A_{i j} & : i = r \end{cases}       | c =  }} {{eqn | ll= \leadsto       | l = \mathbf {E A}       | r = e \paren {\mathbf A}       | c =  }} {{end-eqn}} {{qed|lemma}} === Case $2$ === Let $e$ be the elementary row operation $r_s \to r_s + \lambda r_t$: :$E_{i k} = \begin{cases} \delta_{i k} & : i \ne s \\ \delta_{s k} + \lambda \delta_{t k} & : i = s \end{cases}$ where $\delta$ denotes the Kronecker delta. Then: {{begin-eqn}} {{eqn | l = \sqbrk {E A}_{i j}       | r = \sum_{k \mathop = 1}^m E_{i k} A_{k j}       | c =  }} {{eqn | r = \begin {cases} A_{i j} & : i \ne s \\ A_{i j} + \lambda A_{t j} & : i = s \end {cases}       | c =  }} {{eqn | ll= \leadsto       | l = \mathbf {E A}       | r = e \paren {\mathbf A}       | c =  }} {{end-eqn}} {{qed|lemma}} === Case $3$ === Let $e$ be the elementary row operation $r_s \leftrightarrow r_t$: By Exchange of Rows as Sequence of Other Elementary Row Operations, this elementary row operation can be expressed as: :$e_1 e_2 e_3 e_4 \paren {\mathbf A} = e \paren {\mathbf A}$ where the $e_i$ are elementary row operation of the other two types. For each $e_i$, let $\mathbf E_i = e_i \paren {\mathbf I}$. Then: {{begin-eqn}} {{eqn | l = e \paren {\mathbf A}       | r = e_1 e_2 e_3 e_4 \paren {\mathbf A}       | c = Definition of $e$ }} {{eqn | r = \mathbf E_1 \mathbf E_2 \mathbf E_3 \mathbf E_4 \mathbf A       | c = Cases $1$ and $2$ }} {{eqn | r = \mathbf E_1 \mathbf E_2 \mathbf E_3 e_4 \paren {\mathbf I} \mathbf A }} {{eqn | r = \mathbf E_1 \mathbf E_2 e_3 e_4 \paren {\mathbf I} \mathbf A }} {{eqn | r = \mathbf E_1 e_2 e_3 e_4 \paren {\mathbf I} \mathbf A }} {{eqn | r = e_1 e_2 e_3 e_4 \paren {\mathbf I} \mathbf A }} {{eqn | r = e \paren {\mathbf I} \mathbf A }} {{eqn | r = \mathbf {E A} }} {{end-eqn}} {{qed}}
For ease of notation, define: :$A := \displaystyle \bigcap_{i \mathop = 1}^n A_i$ and let $\tau_A$ denote the included set topology on $S$ by $A$. Let $U \in \tau_A$ be nonempty. Then by definition, $A \subseteq U$. Hence there is a subset $Z \subseteq S$ of $S$, such that $U = A \cup Z$; that is: :$U = \displaystyle \paren {\bigcap_{i \mathop = 1}^n A_i} \cup Z = \bigcap_{i \mathop = 1}^n \paren {A_i \cup Z}$ where the latter equality follows from Union Distributes over Intersection. For every $i \in \closedint 1 n$, one has $A_i \cup Z \in \tau_{A_i}$ by definition of included set topology. Further, by assumption, $\map T {A_i} \le T$, that is $\tau_{A_i} \subseteq \tau$. It follows that $A_i \cup Z \in \tau$ for all $i \in \closedint 1 n$. Hence $U = \displaystyle \bigcap_{i \mathop = 1}^n \paren {A_i \cup Z}$ is in $\tau$, because $\tau$ is a topology. This comes down to: :$\tau_A \subseteq \tau$ Hence by definition of coarser topology: :$\map T {\bigcap A_i} \le T$ {{qed}}
For ease of presentation, let $\breve {\mathbf X}$ be the inverse of $\mathbf X$. We have that $\mathbf A$ can be transformed into $\mathbf I$ by a sequence of elementary row operations. By repeated application of Elementary Row Operations as Matrix Multiplications, we can write this assertion as: {{begin-eqn}} {{eqn | l = \mathbf E_t \mathbf E_{t - 1} \cdots \mathbf E_2 \mathbf E_1 \mathbf A       | r = \mathbf I }} {{end-eqn}} From Elementary Row Matrix is Invertible: :$\mathbf E_1, \dotsc, \mathbf E_t \in \GL {n, \R}$ {{MissingLinks|$\GL {n, \R}$, and explain the significance of this. It's General Linear Group, clearly.}} We can multiply on the left both sides of this equation by: {{begin-eqn}} {{eqn | l = \breve {\mathbf E}_1 \breve {\mathbf E}_2 \cdots \breve {\mathbf E}_{t - 1} \breve {\mathbf E}_t \mathbf E_t \mathbf E_{t - 1} \cdots \mathbf E_2 \mathbf E_1 \mathbf  A        | r = \breve {\mathbf E}_1 \breve {\mathbf E}_2 \cdots \breve {\mathbf E}_{t - 1} \breve {\mathbf E}_t \mathbf I }} {{eqn | ll= \leadsto       | l = \mathbf {II} \cdots \mathbf {IIA}        | r = \breve {\mathbf E}_1 \breve {\mathbf E}_2 \cdots \breve {\mathbf E}_{t - 1} \breve {\mathbf E}_t \mathbf I       | c = definition of inverse }} {{eqn | ll= \leadsto       | l = \mathbf A       | r = \breve {\mathbf E}_1 \breve {\mathbf E}_2 \cdots \breve {\mathbf E}_{t - 1} \breve {\mathbf E}_t       | c = definition of identity }} {{eqn | ll= \leadsto       | l = \breve {\mathbf A}       | r = \breve {\breve {\mathbf E} }_t \breve {\breve {\mathbf E} }_{t - 1} \cdots \breve {\breve {\mathbf E} }_2 \breve {\breve {\mathbf E} }_1       | c = Inverse of Matrix Product, Leibniz's Law }} {{eqn | r = \mathbf E_t \mathbf E_{t - 1} \cdots \mathbf E_2 \mathbf E_1       | c = Inverse of Group Inverse }} {{eqn | r = \mathbf E_t \mathbf E_{t - 1} \cdots \mathbf E_2 \mathbf E_1 \mathbf I       | c = {{Defof|Unit Matrix}} }} {{end-eqn}} By repeated application of Elementary Row Operations as Matrix Multiplications, each $\mathbf E_n$ on the right hand side corresponds to an elementary row operation. Hence the result. {{qed}} {{proofread}} Category:Unit Matrices Category:Inverse Matrices Category:Elementary Row Operations fqmfgxd8fey29torcvzebsnm83fzq1f
For ease of notation, define: :$A := \displaystyle \bigcup_{i \mathop \in I} A_i$ and let $\tau_A$ denote the included set topology on $S$ by $A$. Let $U \in \tau$ be nonempty. As $\map T {A_i}$ is finer than $T$ it follows by definition that: :$\forall i \in I: \tau \subseteq \tau_{A_i}$ Thus: :$\forall i \in I: U \in \tau_{A_i}$ Hence for all $i$ there is a subset $Z_i \subseteq S$ of $S$, such that $U = A_i \cup Z_i$; that is: :$\displaystyle U = \bigcup_{i \mathop \in I} \paren {A_i \cup Z_i} = \paren {\bigcup_{i \mathop \in I} A_i} \cup \paren {\bigcup_{i \mathop \in I} Z_i}$ where the latter equality follows from associativity and commutativity of set union. That is: :$U = A \cup Z$ where: :$\displaystyle Z = \paren {\bigcup_{i \mathop \in I} Z_i}$ Hence $U \in \tau_A$. by definition of the included set topology. This comes down to $\tau \subseteq \tau_A$, and hence $\map T {\bigcup A_i} \ge T$, by definition of a finer topology. {{qed}}
Let $c = \displaystyle \sup_{n \mathop \in \N} \, \left\Vert{T_n}\right\Vert$. By assumption, $c < \infty$. Let $h = \left({h_n}\right)_{n \in \N} \in H$ be arbitrary. Then: {{begin-eqn}} {{eqn|l = \left\Vert{ T h }\right\Vert_H^2      |r = \left\Vert{ \left({T_n h_n}\right)_{n \in \N} }\right\Vert_H^2      |c = Definition of $T$ }} {{eqn|r = \sum_{n \mathop = 1}^\infty \left\Vert{ T_n h_n }\right\Vert_{H_n}^2      |c = Definition of $\left\Vert{\cdot}\right\Vert_H$ }} {{eqn|o = \le      |r = \sum_{n \mathop = 1}^\infty \left\Vert{ T_n }\right\Vert^2 \left\Vert{ h_n }\right\Vert_{H_n}^2 }} {{eqn|o = \le      |r = \sum_{n \mathop = 1}^\infty c^2 \left\Vert{ h_n }\right\Vert_{H_n}^2 }} {{eqn|r = c^2 \sum_{n \mathop = 1}^\infty \left\Vert{ h_n }\right\Vert_{H_n}^2 }} {{eqn|r = c^2 \left\Vert{ h }\right\Vert_H^2 }} {{end-eqn}} {{MissingLinks|results for some steps}} In summary: :$\left\Vert{ T h }\right\Vert_H^2 \le c^2 \left\Vert{ h }\right\Vert_H^2$ It follows that $T$ is bounded. {{qed}} Category:Linear Transformations on Hilbert Spaces t68mo69inf7su8yo7n236b3e1laax81
Since $A$ is normal, we have by Kernel of Linear Transformation is Orthocomplement of Range of Adjoint: Corollary that: :$\ker A = \left({\operatorname{ran} A}\right)^\perp$ and in particular, that: :$\ker A \subseteq \left({\operatorname{ran} A}\right)^\perp$ Now, by Orthocomplement of Subset of Orthocomplement is Superset: :$\operatorname{ran} A \subseteq \left({\ker A}\right)^\perp$ Applying this to the normal operator $A - \lambda$, we find: :$\operatorname{ran} \left({A - \lambda}\right) \subseteq \left({\ker \left({A - \lambda}\right)}\right)^\perp$ {{MissingLinks|Link to result proving $A - \lambda$ normal (not up presently)}} We are now set up to prove that $\ker \left({A - \lambda}\right)$ is a reducing subspace for $A$. Let $x \in \ker \left({A - \lambda}\right)$. Then: {{begin-eqn}} {{eqn|l = A x      |r = \lambda x + \left({A - \lambda}\right) x }} {{eqn|r = \lambda x      |c = Definition of kernel, $x \in \ker \left({A - \lambda}\right)$ }} {{end-eqn}} Therefore, $A \ker \left({A - \lambda}\right) \subseteq \ker \left({A - \lambda}\right)$; that is to say, $\ker \left({A - \lambda}\right)$ is $A$-invariant. Now, let $x \in \left({\ker \left({A - \lambda}\right)}\right)^\perp$. Observe that: :$A x = \lambda x + \left({A - \lambda}\right) x$ Now $\left({A - \lambda}\right) x \in \operatorname{ran} \left({A - \lambda}\right)$, and by our derivation above, this means that: :$\left({A - \lambda}\right) x \in \left({\ker \left({A - \lambda}\right)}\right)^\perp$ In conclusion, since $\left({\ker \left({A - \lambda}\right)}\right)^\perp$ is a linear subspace of $H$, it follows that: :$\lambda x + \left({A - \lambda}\right) x \in \left({\ker \left({A - \lambda}\right)}\right)^\perp$ as desired. Hence both $\ker \left({A - \lambda}\right)$ and $\left({\ker \left({A - \lambda}\right)}\right)^\perp$ have been shown to be $A$-invariant subspaces of $H$. That is, $\ker \left({A - \lambda}\right)$ is a reducing subspace for $A$. {{qed}}
{{improve}} Let $\mathcal V$ be an inner product space. {{explain|The theorem calls for a Hilbert space. What is actually required?}} Let $T: \mathcal V \to \mathcal V$ be a normal linear operator. Requisite knowledge: $T^*$ is the adjoint of $T$ and is defined by the fact that for any $u, w \in \mathcal V$, we have :$\left\langle{T u, w}\right\rangle = \left\langle{T^* w}\right\rangle$ It is important to note the existence and uniqueness of adjoint operators. {{explain|Link to the required proofs and/or definitions as given above.}} '''Claim''': We know that for $v \in \mathcal V$: :$T v = \lambda v \iff T^* v = \overline \lambda v$ This is true because for all normal operators, by definition: :$T^* T = T T*$ and so: {{begin-eqn}} {{eqn | l = \left\Vert{T v}\right\Vert^2       | r = \left\langle{T v, T v}\right\rangle       | c =  }} {{eqn | r = \left\langle{T^* T v, v}\right\rangle       | c =  }} {{eqn | r = \left\langle{T T^* v, v}\right\rangle       | c =  }} {{eqn | r = \left\langle{T^* v, T^*v}\right\rangle       | c =  }} {{eqn | r = \left\Vert{T^* v}\right\Vert^2       | c =  }} {{end-eqn}} Since for normal $T$, $\left({T - \lambda I}\right)$ is normal, we have: {{explain|Link to a page demonstrating the above}} {{begin-eqn}} {{eqn | l = T v       | r = \lambda v       | c =  }} {{eqn | ll= \iff        | l = \left\Vert{\left({T - \lambda I}\right) v}\right\Vert       | r = 0       | c =  }} {{eqn | ll= \iff        | l = \left\Vert{\left({T - \lambda I}\right)^* v}\right\Vert       | r = 0       | c =  }} {{eqn | ll= \iff        | l = \left\Vert{T^* v - \overline \lambda v}\right\Vert       | r = 0       | c =  }} {{eqn | ll= \iff        | l = T^* v       | r = \overline \lambda v       | c =  }} {{end-eqn}} Now, if $T v_1 = \lambda_1 v_1$ and $T v_2 = \lambda_2 v_2$, where $\lambda_1 \ne \lambda_2$ and $v_1, v_2$ are eigenvectors (i.e. $v_1, v_2 \ne \vec 0$), we have: {{begin-eqn}} {{eqn | l = \lambda_1 \left\langle{v_1, v_2}\right\rangle       | r = \left\langle{\lambda_1 v_1, v_2}\right\rangle       | c =  }} {{eqn | r = \left\langle{T v_1, v_2}\right\rangle       | c =  }} {{eqn | r = \left\langle{v_1, T^* v_2}\right\rangle       | c =  }} {{eqn | r = \left\langle{v_1, \overline{\lambda_2} v_2}\right\rangle       | c =  }} {{eqn | r = \lambda_2 \left\langle{v_1, v_2}\right\rangle       | c =  }} {{end-eqn}} Since $\lambda_1 \ne \lambda_2$, this is only possible if $\left\langle{v_1, v_2}\right\rangle = 0$, which means the eigenvectors of our normal operator are orthogonal. {{qed}}
Let $\lambda$ be an eigenvalue of $A$. Let $v \in H$ be an eigenvector for $\lambda$; i.e. $Av = \lambda v$. Now compute: {{begin-eqn}} {{eqn|l = \lambda \left\langle{v, v}\right\rangle      |r = \left\langle{\lambda v, v}\right\rangle      |c = Property $(2)$ of an inner product }} {{eqn|r = \left\langle{Av, v}\right\rangle      |c = $v$ is an eigenvector }} {{eqn|r = \left\langle{v, Av}\right\rangle      |c = $A$ is self-adjoint }} {{eqn|r = \overline{\left\langle{Av, v}\right\rangle}      |c = Property $(1)$ of an inner product }} {{eqn|r = \overline{\left\langle{\lambda v, v}\right\rangle}      |c = $v$ is an eigenvector }} {{eqn|r = \bar \lambda \left\langle{v, v}\right\rangle      |c = Properties $(2)$, $(4)$ of an inner product  }} {{end-eqn}} Now $v$, being an eigenvector, is non-zero. By property $(5)$ of an inner product, this implies $\left\langle{v, v}\right\rangle \ne 0$. Thence, dividing out $\left\langle{v, v}\right\rangle$, obtain $\lambda = \bar \lambda$. From Complex Number equals Conjugate iff Wholly Real, $\lambda \in \R$. {{qed}}
From Hartogs' lemma there exists an ordinal $\alpha$ such that there is no injection from $\alpha$ to $S$. We also have that Ordinals are Well-Ordered. It follows from the definition of well-ordering that there exists a smallest such ordinal. Hence the result. {{qed}}
Follows immediately from Cardinal Equal to Collection of All Dominated Ordinals. The collection of all dominated ordinals: :$\set {y \in \On: y \preccurlyeq S}$ is the Hartogs number of $S$. {{qed}}
Proof by Counterexample: Consider the algebraic structure $\left({S, \circ}\right)$ consisting of: :The set $S = \left\{{a, b, e}\right\}$ :The binary operation $\circ$ whose Cayley table is given as follows: :$\begin{array}{c|cccc} \circ & e & a & b \\ \hline e & e & a & b \\ a & a & e & e \\ b & b & e & e \\ \end{array}$ By inspection, we see that $e$ is the identity element of $\left({S, \circ}\right)$. We also note that: {{begin-eqn}} {{eqn | l = \left({a \circ a}\right) \circ b       | r = e \circ b       | c =  }} {{eqn | r = b       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a \circ \left({a \circ b}\right)       | r = a \circ e       | c =  }} {{eqn | r = a       | c =  }} {{end-eqn}} and so $\circ$ is not associative. Note further that: {{begin-eqn}} {{eqn | l = a \circ b       | r = e       | c =  }} {{eqn | r = b \circ a       | c =  }} {{end-eqn}} and also: {{begin-eqn}} {{eqn | l = a \circ a       | r = e       | c =  }} {{end-eqn}} So both $a$ and $b$ are inverses of $a$. Hence the result. {{qed}}
Suppose: : $\exists g_1: T \to S: g_1 \circ f = I_S$ : $\exists g_2: T \to S: f \circ g_2 = I_T$ From Injection iff Left Inverse, it follows that $f$ is an injection. From Surjection iff Right Inverse, it follows that $f$ is a surjection. So $f$ is both an injection and a surjection and, by definition, therefore also a bijection.
Suppose: : $\exists g_1: T \to S: g_1 \circ f = I_S$ : $\exists g_2: T \to S: f \circ g_2 = I_T$ We have that the Identity Mapping is Bijection, thus $I_S$ and $I_T$ are both bijections. From Injection if Composite is Injection, if $I_S = g_1 \circ f$ is an injection, then so is $f$. From Surjection if Composite is Surjection, if $I_T = f \circ g_2$ is a surjection, then so is $f$. So $f$ is both an injection and a surjection and, by definition, therefore also a bijection. {{qed}}
{{begin-eqn}} {{eqn | l = g_2       | r = I_S \circ g_2       | c = {{Defof|Identity Mapping}} }} {{eqn | r = \paren {g_1 \circ f} \circ g_2       | c = by hypothesis }} {{eqn | r = g_1 \circ \paren {f \circ g_2}       | c = Composition of Mappings is Associative }} {{eqn | r = g_1 \circ I_T       | c = by hypothesis }} {{eqn | r = g_1       | c = {{Defof|Identity Mapping}} }} {{end-eqn}} From Left and Right Inverse Mappings Implies Bijection it follows that $f$ is a bijection. It follows from Composite of Bijection with Inverse is Identity Mapping that $g_1 = g_2 = f^{-1}$. {{qed}}
From Left and Right Inverse Mappings Implies Bijection, we have that $f$ is a bijection. Let $y \in T$ and let $x = f^{-1} \left({y}\right)$. Such an $x$ exists because $f$ is surjective, and unique within $S$ as $f$ is injective. Then $y = f \left({x}\right)$ and so: {{begin-eqn}} {{eqn | l = f^{-1} \left({y}\right)       | r = x       | c =  }} {{eqn | r = I_S \left({x}\right)       | c = $I_S$ is the identity mapping on $S$ }} {{eqn | r = g_1 \circ f \left({x}\right)       | c = by hypothesis }} {{eqn | r = g_1 \left({y}\right)       | c = from above: $y = f \left({x}\right)$ }} {{end-eqn}} So $f^{-1} = g_1$. Also: {{begin-eqn}} {{eqn | l = f \left({x}\right)       | r = y       | c =  }} {{eqn | r = I_T \left({y}\right)       | c = $I_T$ is the identity mapping on $T$ }} {{eqn | r = f \circ g_2 \left({y}\right)       | c = by hypothesis }} {{eqn | ll= \implies       | l = f^{-1} \left({y}\right)       | r = x       | c = as $f$ is injective }} {{eqn | r = g_2 \left({y}\right)       | c =  }} {{end-eqn}} So $f^{-1} = g_2$. {{qed}}
Because Composition of Mappings is Associative, brackets do not need to be used. {{begin-eqn}} {{eqn | n = 1       | l = g_1 \circ f       | r = I_S       | c =  }} {{eqn | ll= \implies       | l = g_1 \circ f \circ g_2       | r = I_S \circ g_2       | c =  }} {{eqn | r = g_2       | c = {{Defof|Identity Mapping}} }} {{end-eqn}} {{begin-eqn}} {{eqn | n = 2       | l = f \circ g_2       | r = I_T       | c =  }} {{eqn | ll= \implies       | l = g_1 \circ f \circ g_2       | r = g_1 \circ I_T       | c =  }} {{eqn | r = g_1       | c = {{Defof|Identity Mapping}} }} {{end-eqn}} Thus $g_1 = g_2$. Now suppose there exists $g_3: T \to S: g_3 \circ f = I_S$. By the same argument as above, $g_3 = g_2$. This means that $g_1 (= g_3)$ is the only left inverse of $f$. Similarly, suppose there exists $g_4: T \to S: f \circ g_4 = I_T$. By the same argument as above, $g_4 = g_1$. This means that $g_2 (= g_4)$ is the only right inverse of $f$. So $g_1 = g_2 = g_3 = g_4$ are all the same. By Composite of Bijection with Inverse is Identity Mapping, it follows that this unique mapping is the inverse $f^{-1}$. {{qed}}
By definition of composition of mappings: :$f_2 \circ f_1 = \left\{{\left({x, z}\right) \in S_1 \times S_3: \exists y \in S_2: \left({x, y}\right) \in f_1 \land \left({y, z}\right) \in f_2}\right\}$ Let $\mathcal R_1 \subseteq S_1 \times T_1$ and $\mathcal R_2 \subseteq S_2 \times T_2$ be relations. The composite of $\mathcal R_1$ and $\mathcal R_2$ is defined as: :$\mathcal R_2 \circ \mathcal R_1 := \left\{{\left({x, z}\right) \in S_1 \times T_2: \exists y \in S_2 \cap T_1: \left({x, y}\right) \in \mathcal R_1 \land \left({y, z}\right) \in \mathcal R_2}\right\}$ When $T_1 = S_2$ and $T_2 = S_3$, this becomes: :$\mathcal R_2 \circ \mathcal R_1 := \left\{{\left({x, z}\right) \in S_1 \times T_3: \exists y \in S_2: \left({x, y}\right) \in \mathcal R_1 \land \left({y, z}\right) \in \mathcal R_2}\right\}$ The definitions can be seen to be identical. {{qed}}