Proposition 1.1. Suppose a relation $f\subset A\times B$ satisfies one of inclusions

$$\begin{array}{llllllll}\hfill & \left(1\right)f{f}^{-1}\subset 1_B,& \hfill & \left(2\right)f^{-1}f\supset 1_A,& \hfill & & \hfill & \\ \hfill & \left(3\right)f^{-1}f\subset 1_A,& \hfill & \left(4\right)f{f}^{-1}\supset 1_B.& \hfill & & \hfill & \end{array}$$

If inclusion $\left(1\right)$ holds ( inclusion $\left(2\right)$, respectively), then for any element $a\in A$ the image $f\left(a\right)$ contains at most (at least) one element from set $B$. If inclusion $\left(3\right)$ holds (inclusion $\left(4\right)$, respectively), then for any element $b\in B$ the original $f^{-1}\left(b\right)$ contains at most (at least) one element from $A$.

Definition 1.1. The relation $f\subset A\times B$ is said to be

– a function from set $A$ to set $B$, if inclusion (1) holds;

– a map from set $A$ to set $B$, if inclusions (1) and (2) hold;

– an injection from set $A$ to set $B$, if inclusions (1), (2) and (3) hold;

– a surjection from set $A$ onto set $B$, if inclusions (1), (2) and (4) hold;

– a bijection between sets $A$ and $B$, if inclusions (1), (2), (3) and (4) hold;

– a multi-valued map from set $A$ to set $B$, if inclusion (2) holds.

Theorem 2.1. a) Suppose the maps $g$ and $h$ are given. Let the inclusion $\left(5\right)$ hold. Then there exists a map $f$ such that $h=gf$. Moreover, the map $f$ is obtained by restriction of the multi-valued map $g^{-1}h$.

b) Suppose the maps $f$ and $h$ are given. Let the inclusion $\left(6\right)$ hold. Then there exists a map $g$ such that $h=gf$. Moreover, the map $g$ is obtained by extension of the function $h{f}^{-1}$.

Remark. Notice, that we don’t use elements in proof of Theorem 2.1. But if we use elements, then in the case a) the inclusion (5) is equivalent to the inclusion $h\left(A\right)\subset g\left(B\right)$. We construct the map $f:A\to B$ so that for element $a\in A$ the image $f\left(a\right)\in B$ has to satisfy the condition $g\left(f\left(a\right)\right)\in h\left(A\right)$. Because of $h\left(A\right)\subset g\left(B\right)$ it is possible that the condition $g\left(f\left(a\right)\right)\in h\left(A\right)$ holds for each element $a\in A$. In the case b) the set $A$ has two equivalence relations $f^{-1}f$ and $h^{-1}h$. Thus we have two partitions of $A$ into equivalence classes. From (6) it follows that the first partition is a refinement of the second one. For $b\in f\left(A\right)$ we have $g\left(b\right)=h{f}^{-1}\left(b\right)$ since $h{f}^{-1}$ is a function. For $b\in B$ outside $f\left(A\right)$ we choose $g\left(b\right)$ arbitrary.

However, it is preferable to use the inclusions (1)–(6) instead of using elements of the sets. As they say, these inclusions formalize proofs.

Consecuence 1. Suppose the inclusion (5) holds and the map $g$ is an injection. Then the relation $g^{-1}h$ is a map: $g^{-1}h{\left(g^{-1}h\right)}^{-1}=g^{-1}h{h}^{-1}g\subset g^{-1}g=1_B$ (besides (2) inclusion (1) also holds for this relation), and the map $f$ is defined uniquely by $f=g^{-1}h$.

Consecuence 2. Suppose the inclusion (6) holds and the map $f$ is a surjection. Then the relation $h{f}^{-1}$ is a map: ${\left(h{f}^{-1}\right)}^{-1}h{f}^{-1}=f{h}^{-1}h{f}^{-1}\supset f{f}^{-1}=1_B$ (besides (1) inclusion (2) also holds for this relation), and the map $g$ is defined uniquely by $g=h{f}^{-1}$.

Consecuence 3. If $g$ is a surjection, $g{g}^{-1}=1_C$, then inclusion (5) is valid automatically and there always exists the map $f$ such that $h=gf$.

Consecuence 4. If $f$ is an injection, $f^{-1}f=1_A$, then inclusion (6) is valid automatically and there always exists the map $g$ such that $h=gf$.

Consecuence 5. For a composition $h=gf$ with $h$ injective it follows that $f$ is also injective:

Similarly, if $h$ is a surjection, then $g$ is a surjection too:

In particular, if $gf=1_A$, then the map $f$ is always injection and the map $g$ is always surjection. In this case $f$ is called a section of the surjection $g$, while $g$ is called a retraction of the injection $f$.

Definition 2.1. If in the composition $h=gf$ the map $f$ is a surjection and $g$ is an injection, the commutative diagram

is said to represent a canonical representation of the map $h$.

Theorem 2.2. For any map $h:A\to C$ the canonical representation is defined up to natural bijection. It means if $h$ has two canonical representations $h=g_1{f}_1$ and $h=g_2{f}_2$, see diagram below, then there exists the unique bijection $\beta :B_1\to B_2$ such that the entire diagram

is commutative, i.e., $f_2=\beta f_1$ and $g_1=g_2\beta$.

Theorem 2.3. Suppose we have a prism diagram with faces I, II and III:

where the face I is a commutative square, i.e., $\gamma h=h\prime \alpha$. The maps $h$ and $h\prime$ have canonical representations $h=gf$ and $h\prime =g\prime f\prime$, respectively. Then there exists the unique map $\beta$ such that the faces II and III are commutative.

Definition 2.2. If in the prism diagram from Theorem 3 the faces I, II and III are commutative, a triple $\left(\alpha ,\beta ,\gamma \right)$ is said to define the morphism of two diagrams $ABC$ and $A\prime B\prime C\prime$. A morphism is a monomorphism (epimorphism) of diagrams, if $\alpha$, $\beta$ and $\gamma$ are injections (surjections, respectively). A morphism, which is both epi- and monomorphism, is called an isomorphism of diagrams. A canonical representation of morphism of diagrams $ABC$ and $A\prime B\prime C\prime$ is it presentation in composition of epimorphism and monomorphism.

Theorem 2.4. The canonical representation of morphism of diagrams $ABC$ and $A\prime B\prime C\prime$ exists up to isomorphism. It means that there exists a commutative diagram $UVW$ such that the entire diagram

is commutative and the diagram $UVW$ is defined up to isomorphism.

Theorem 2.5. For any commutative square $ACC\prime A\prime$ there exists a decomposition on blocks:

where the northwest block $ABVU$ consists of surjections, while the southeast block $VWC\prime B\prime$ consists of injections. This decomposition is defined up to isomorphism.

Theorem 2.6. Let $h$ be a map from set $A$ to itself:

Let us form the sequence of maps

where $h_1=f_1{g}_1:B_1\to B_1$ is defined after canonical representation $h=g_1{f}_1$, and $h_k=f_k{g}_k:B_k\to B_k$ after canonical representation $h_{k-1}=g_k{f}_k,k=2,3,...$ If $A$ is a finite set, then there exist the minimal integers $m$ and $n$, such that

It means, that after $n$ steps the iteration $h^m$ repeats if and only if $h_m:B_m\to B_m$ is a bijection and ${\left(h_m\right)}^n$ is identical map.

Theorem 3.1. Suppose $p$ is the rank of matrix $H$. Then the matrix $H$ can be represented as a product $H=GF$, where $F$ is a surjective $p\times n$-matrix and $G$ is an injective $m\times p$-matrix. Therefore, for any other product of surjective $p\times n$-matrix $F\prime$ and injective $m\times p$-matrix $G\prime$ such that $H=G\prime F\prime$, there exists a non-degenerated $p\times p$-matrix $Q$ such that $G\prime =G{Q}^{-1}$, $F\prime =QF$ and $H=\left(G{Q}^{-1}\right)\left(QF\right)$.

Example. Let

be a map from set $A=\left\{1,2,3,4,5\right\}$ to itself. Let us represent the map $h$ as a square matrix $H=\left(h_{ij}\right)$ by

Corresponding to the scheme from Theorem 2.6 the equality of iterations $h^4=h^2$ can be written

$$\begin{array}{llll}\hfill h^4& =g_1{g}_2{h}_2^2{f}_2{f}_1,& \hfill & \\ \hfill H^4& =G_1{G}_2{H}_2^2{F}_2{F}_1,& \hfill & \end{array}$$

Each matrix from the last equality corresponds to a linear map: endomorphism $H:{\mathbb{ℝ}}^5\to {\mathbb{ℝ}}^5$, two epimorphisms $F_1:{\mathbb{ℝ}}^5\to {\mathbb{ℝ}}^3$, $F_2:{\mathbb{ℝ}}^3\to {\mathbb{ℝ}}^2$, automorphism $H_2:{\mathbb{ℝ}}^2\to {\mathbb{ℝ}}^2$ and two monomorphisms $G_2:{\mathbb{ℝ}}^2\to {\mathbb{ℝ}}^3$, $G_1:{\mathbb{ℝ}}^3\to {\mathbb{ℝ}}^5$ (in this paper, matrices and corresponding linear maps are denoted by the same symbol). After two iterations we have

Let $\left\{x^i\right\}$ be the coordinates of space ${\mathbb{ℝ}}^5$ according to the usual basis $\left\{e_i\right\},i=1,...,5$. Then the kernel Ker$H$ is a two-dimensional coordinate plane $x^1=x^2=x^3=0$, the kernel Ker$H^2$ is a three-dimensional coordinate space $x^1=x^2=0$, the image Im$H$ is a three-dimensional hyperplane spanned by vectors $\left\{e_2+e_3,e_1+e_5,e_4\right\}$ and Im$H^2$ is a two-dimensional plane spanned by vectors $\left\{e_2+e_3,e_1+e_4+e_5\right\}$.The second iteration of the automorphism $H_2$ is the identical map of the plane Im$H^2$. Thus, from Theorem 2.6 it follows that

where $E_2$ is a unit matrix of order 2.

This result is useful in the next situation. Let us consider the matrix $H$ as an element of Lie algebra $gl\left(5,\mathbb{ℝ}\right)$. Then the exponential $e^{tH}$ is a one-parameter subgroup of Lie group $GL\left(5,\mathbb{ℝ}\right)$. Since $H^4=H^2$ we have

The linear vector field corresponding to the matrix $H$ is

Note that all vectors in $X$ are parallel to Im$H$. The field $X$ has a canonical parameter

and four invariants

The exponential law $U\prime =HU\Rightarrow U_t=e^{tH}U$ defines a flow $a_t:U\mapsto U_t$ (see $e^{tH}$),

generated by vector field $X$. In particular, the points, represented by basis vectors, move along trajectories

$$\begin{array}{llll}\hfill {\left(e_1\right)}_t& =-e_4-e_5+\left(e_1+e_4+e_5\right)cosht+\left(e_2+e_3\right)sinht,& \hfill & \\ \hfill {\left(e_2\right)}_t& =-e_3-e_{4}t+\left(e_2+e_3\right)cosht+\left(e_1+e_4+e_5\right)sinht,& \hfill & \\ \hfill {\left(e_3\right)}_t& =e_3+e_{4}t,& \hfill & \\ \hfill {\left(e_4\right)}_t& =e_4,& \hfill & \\ \hfill {\left(e_5\right)}_t& =e_5.& \hfill & \end{array}$$

It means that under the action of $a_t$ the points $U\in$Ker$H$ are fixed (see ${\left(e_4\right)}_t$ and ${\left(e_5\right)}_t$), the points $U\in$ Ker$H^2\setminus$Ker$H$ move along the straight lines (see ${\left(e_3\right)}_t$) and the points $U\in {\mathbb{ℝ}}^2\setminus$Ker$H^2$ participate in a hyperbolic rotation in the plane Im$H^2$ (see ${\left(e_1\right)}_t$ and ${\left(e_2\right)}_t$). Note that the hyperbolic rotation in the plane Im$H^2$ can be described using the basis vectors:

$$\begin{array}{lll}\hfill {\left(e_1+e_4+e_5\right)}_t=\left(e_1+e_4+e_5\right)cosht+\left(e_2+e_3\right)sinht,& & \hfill \\ \hfill {\left(e_2+e_3\right)}_t=\left(e_2+e_3\right)cosht+\left(e_1+e_4+e_5\right)sinht.& & \hfill \end{array}$$

The coordinate transformation $\left(x^1,x^2,x^3,x^4,x^5\right)\mapsto \left(s,I_1,I_2,I_3,I_4\right)$ transforms the basis $\left\{e_i\right\}$ into another invariant basis: a coframe $\left(ds,d{I}_1,d{I}_2,d{I}_3,d{I}_4\right)$ and a frame

$$\begin{array}{llll}\hfill \frac{\partial }{\partial s}& =X,& \hfill & \\ \hfill \frac{\partial }{\partial I_1}& =-\frac{\partial }{\partial x^5},& \hfill & \\ \hfill \frac{\partial }{\partial I_2}& =-\frac{\partial }{\partial x^3}-s\frac{\partial }{\partial x^4},& \hfill & \\ \hfill \frac{\partial }{\partial I_3}& =x^1\left(\frac{\partial }{\partial x^1}+\frac{\partial }{\partial x^5}\right)+x^2\left(\frac{\partial }{\partial x^2}+\frac{\partial }{\partial x^3}+\frac{\partial }{\partial x^4}\right),& \hfill & \\ \hfill \frac{\partial }{\partial I_4}& =-x^1\left(\frac{\partial }{\partial x^2}+\frac{\partial }{\partial x^3}\right)-x^2\left(\frac{\partial }{\partial x^1}+\frac{\partial }{\partial x^4}+\frac{\partial }{\partial x^5}\right).& \hfill & \end{array}$$

This transformation follows from Jacobi matrix and its inverse:

$$\begin{array}{c}{\left(\begin{array}{ccccc}\hfill \frac{1}{I_2}\hfill & \hfill -\frac{s}{I_2}\hfill & \hfill \frac{s}{I_2}\hfill & \hfill -\frac{1}{I_2}\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill \frac{x^1}{{\left(x^1\right)}^2-{\left(x^2\right)}^2}\hfill & \hfill -\frac{x^2}{{\left(x^1\right)}^2-{\left(x^2\right)}^2}\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill \frac{1}{I_2}+\frac{x^2}{{\left(x^1\right)}^2-{\left(x^2\right)}^2}\hfill & \hfill -\frac{s}{I_2}-\frac{x^1}{{\left(x^1\right)}^2-{\left(x^2\right)}^2}\hfill & \hfill \frac{s}{I_2}\hfill & \hfill -\frac{1}{I_2}\hfill & \hfill 0\hfill \\ \hfill \hfill \end{array}\right)}^{-1}=\\ =\left(\begin{array}{ccccc}\hfill x^2\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill x^1\hfill & \hfill -x^2\hfill \\ \hfill x^1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill x^2\hfill & \hfill -x^1\hfill \\ \hfill x^1\hfill & \hfill 0\hfill & \hfill -1\hfill & \hfill x^2\hfill & \hfill -x^1\hfill \\ \hfill x^3\hfill & \hfill 0\hfill & \hfill -s\hfill & \hfill x^2\hfill & \hfill -x^2\hfill \\ \hfill x^2\hfill & \hfill -1\hfill & \hfill 0\hfill & \hfill x^1\hfill & \hfill -x^2\hfill \\ \hfill \hfill \end{array}\right).\end{array}$$

The operators $\left(\frac{\partial }{\partial s},\frac{\partial }{\partial I_1},\frac{\partial }{\partial I_2},\frac{\partial }{\partial I_3},\frac{\partial }{\partial I_4}\right)$ commute with vector field $X$. Hence, these operators are infinitesimal symmetries of $X$. From $\frac{\partial }{\partial s}=X$ it follows that the flow generated by $X$ becomes a group of translations $s\mapsto s+t$. In the coordinates $\left(s,I_1,I_2,I_3,I_4\right)$ the trajectories are the lines $s$ and the classification of these trajectories makes no sense.

We have done the following seven steps:

a) took a map $h$ from set $A$ (consists of 5 elements) to itself;

b) represented the map $h$ as $5\times 5$-matrix $H$ with the rank 3;

c) considered the matrix $H$ as an element of Lie algebra $gl\left(5,\mathbb{ℝ}\right)$;

d) found the one-parameter subgroup $e^{tH}$ of Lie group $GL\left(5,\mathbb{ℝ}\right)$;

e) in space ${\mathbb{ℝ}}^5$ the exponential $e^{tH}$ defines a flow $a_t$ generated by corresponding vector field $X$;

f) the classification of the trajectories depends on kernels Ker$H$ and Ker$H^2$;

g) in curvilinear coordinates such as the invariant coordinates $\left(s,I_1,I_2,I_3,I_4\right)$ the flow simplifies, but the classification makes no sense.

The steps b)–g) can be applied to any square matrix with an arbitrary rank.