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Every endomorphism of the additive finite cyclic group $C_n=\mathbb{Z}/n\mathbb{Z}$, for a positive integer $n$, has the form $f_a(k)=ak(\mathrm{mod}n)$, determined by $a=f_a(1)$. This is the self-map case of homomorphism from a finite cyclic group: repeated addition forces the value at every element. Conversely, each choice of $a$ respects addition and gives a well-defined map on congruence classes.

Let $d=\gcd (n,a)$, the greatest common divisor. The image of a group homomorphism is the cyclic subgroup generated by $a$, with size $n/d$ by order of an element of a finite cyclic group. The kernel of a group homomorphism isIndeed, $ak=0(\mathrm{mod}n)$ is equivalent to $n/d$ dividing $k$, after cancelling $d$ and using coprimality. Consequently $f_a$ is a group automorphism exactly when $a$ and $n$ are coprime. In that case a modular inverse $b$ of $a$ gives $f_a^{-1}=f_b$. This connects reversible self-maps directly to the generators of a finite cyclic group.^[ref]

Exterior product of differential forms combined with endomorphism composition is associative and generally noncommutative. For an endomorphism-valued one-form $a$, $(a\wedge a)(u,v)=[a(u),a(v)]$. The endomorphism bundle connection extends as a graded derivation of this algebra.

For endomorphisms $S,T$ of a finite-dimensional space of dimension $d$, $r(S+T)\le r(S)+r(T)$ and $n(ST)\le n(S)+n(T)$. If both equalities hold, the first gives $r(S)+r(T)\le d$, while the second gives $2d-r(S)-r(T)\le d$. Thus the rank sum is $d$, $S+T$ is an isomorphism, and $ST=0$. Conversely, $ST=0$ implies $\mathrm{im}T\subseteq \ker S$ and hence rank sum at most $d$, whereas invertibility of $S+T$ gives rank sum at least $d$. This proves both equalities.

An endomorphism of $F_n$ has image generated by at most $n$ elements. If this image has finite index $d$, the Nielsen–Schreier formula gives rank $1+d(n-1)$. Its abelianization requires at least that many generators. For $n\ge 2$, comparison with $n$ forces $d=1$.

// Target: geometry-and-topology.bigb

Let $G$ be a finitely generated group and $\theta :G\to G$ a surjective endomorphism. Pullback by $\theta$ permutes the finite set of normal subgroups of any prescribed finite index. For each such $N$, some iterate satisfies ${\theta }^{-m}(N)=N$, so $\ker \theta \subseteq \ker {\theta }^m\subseteq N$. Intersecting proves the inclusion. Consequently every finitely generated residually finite group is a Hopfian group.

The Lefschetz operator commutes with the bundle-valued Dolbeault Laplacian on the entire smooth form space exactly when the Chern connection is flat. The Lefschetz-Dolbeault curvature commutator gives one direction. Conversely test its curvature-wedge operator on degree-zero sections with arbitrary prescribed values at any point to force every vector-bundle curvature endomorphism to vanish.

The endomorphisms of a vector space form a Lie algebra under the commutator $[A,B]=AB-BA$. It is the Lie algebra of the general linear group; the Jacobi identity follows by expanding associative products.

A group automorphism is a bijective group homomorphism $\alpha :G\to G$. It preserves the group operation, the identity element, and inverse elements; its inverse is another group homomorphism. Thus it is a group isomorphism with identical source and target. These maps form the automorphism group $\mathrm{Aut}(G)$ under function composition.^[ref]

For the additive finite cyclic group $C_6$, take $f_5(k)=5k(\mathrm{mod}6)$. Its output list on $0,1,2,3,4,5$ is $0,5,4,3,2,1$, a permutation of all six elements. It preserves addition because $5(a+b)=5a+5b(\mathrm{mod}6)$, and applying it twice gives $25k=k(\mathrm{mod}6)$. Hence it is a group automorphism and its own inverse function. Compare the collapsing $f_2$ in the figure under endomorphism. The general criterion under endomorphism of a finite cyclic group says exactly when such a map is reversible: its multiplier must be coprime to the size of the finite cyclic group.

A permutation need not be a group automorphism, even if it fixes the identity element. Let $\pi$ exchange $1$ and $2$ in $C_6$ and fix $0,3,4,5$. It is a bijection, and $\pi (0)=0$, butThus it fails to preserve addition and is not even a group homomorphism. Reversibility alone does not preserve the group operation.

A group automorphism carries each cyclic subgroup onto the cyclic subgroup generated by the image of its defining element, preserving the order of a group element. In particular, it carries a generator of a group to another generator of a group. For $C_6$, the existing calculation of generators of a finite cyclic group leaves only the images $1$ and $5$ for $1$, so its automorphism group consists exactly of the identity function and $f_5$. This turns the problem of compatible relabeling into a much smaller choice than enumerating arbitrary permutations.

A group isomorphism is a bijective group homomorphism $\phi :G\to H$. It matches the entire group operation: $\phi (ab)=\phi (a)\phi (b)$, with each product taken in its own group. Its inverse is automatically a group homomorphism. A group automorphism is the special case $H=G$.^[ref]

The target is part of the claim. The doubling example under isomorphism is a group isomorphism $d:\mathbb{Z}\to 2\mathbb{Z}$, between the additive integers and the additive even integers. Exactly the same rule $d(k)=2k$, declared as $d:\mathbb{Z}\to \mathbb{Z}$, is an injective endomorphism that misses $1$ and every other odd integer, so is not a group automorphism. Changing the declared target changes surjectivity, even though it does not change the formula or its compatibility with addition.

An isomorphism lets a calculation in one model solve the same problem in another. The example under group identifies addition modulo four with function composition of rotations of a square via $k\mapsto r^k$. Through this group isomorphism, adding exponents corresponds exactly to composing motions. More generally, a group isomorphism preserves order of a group element, carries subgroups to subgroups, and carries a generating set of a group to a generating set of a group. Having equally many elements is insufficient: the permutation in group automorphism already fails compatibility on two inputs.

Why does the inverse preserve the group operation? Write $u=\phi (a)$ and $v=\phi (b)$, possible because of surjectivity. ThenThis proves the inverse condition directly. Analogous calculations work for each operation of a ring and for addition and scalar multiplication in a vector space. The argument relies on preserving those operations; it supplies no general continuity claim for the inverse of a continuous bijection between topological spaces.

On the torus ${\mathbb{R}}^d/{\mathbb{Z}}^d$, an integer matrix $A$ defines an endomorphism. The coordinate circles give a basis of the first homology group, and the induced map on that group is $A$. The Künneth theorem identifies higher integral homology with the exterior powers of ${\mathbb{Z}}^d$, so the degree-$q$ map is ${\bigwedge }^{q}A$. In top degree it is multiplication by $\det A$. The map is a homeomorphism precisely when $A$ is a unimodular matrix; a merely nonsingular integer matrix need not have an integer inverse.

The Kronecker quiver has two vertices and two parallel arrows from the first to the second. For a representation with arrows $I,N$, an endomorphism consists of the same matrix at both vertices and must commute with $N$. If $N$ is a single Nilpotent Jordan block, this centralizer is $k[N]$, yielding an indecomposable non-brick when the block size exceeds one.

A Lubin–Tate formal group associated with a Lubin–Tate series $f$ is the unique one-dimensional commutative formal group law ${\mathcal{F}}_f$ over the valuation ring ${\mathcal{O}}_K$ for which $f$ is an endomorphism with linear coefficient the chosen uniformizer $\pi$. The Lubin–Tate functional equation lemma constructs the law and the endomorphisms $[a{]}_f$ for every $a\in {\mathcal{O}}_K$, with linear term $aX$. Their addition is given by the law, and their multiplication by composition. Thus the valuation ring acts on the group; $[\pi {]}_f=f$.

Choose an ordered basis $(e_1,\dots ,e_n)$ of a finite-dimensional vector space. The matrix of a bilinear form records all pairings of basis vectors, with $A_{ij}=B(e_i,e_j)$. Writing $v={\sum }_i{v}_i{e}_i$ and $w={\sum }_j{w}_j{e}_j$, separate linearity giveswhere the last expression uses the coordinate vectors of $v,w$. Conversely, every matrix $A$ defines a bilinear form by this formula.^[ref]

For a concrete bilinear form on ${\mathbb{R}}^2$, takeThen $B((1,0),(0,1))=2$ but $B((0,1),(1,0))=0$: a bilinear form need not be a symmetric bilinear form. For $v=(2,1{)}^T$ and $w=(-1,3{)}^T$, first compute $Aw=(5,3{)}^T$, then $v^{T}Aw=2\cdot 5+1\cdot 3=13$. Reversing the inputs gives $Av=(4,1{)}^T$ and $w^{T}Av=-4+3=-1$.

Under change of basis, let the columns of an invertible matrix $P$ be the new basis vectors in old coordinates. Thus $v=P{v}^{\prime }$ and $w=P{w}^{\prime }$, soThis is matrix congruence, rather than the matrix similarity $P^{-1}AP$ for an endomorphism. Both inputs have changed coordinates, which explains the two factors of $P$. For the displayed $A$ and $P=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$, direct multiplication gives $A^{\prime }=\left(\begin{array}{cc}1& 3\\ 1& 4\end{array}\right)$; its $(1,2)$ entry is $B(e_1,e_1+e_2)=3$.^[ref]

Composition of endomorphisms and wedge product of differential forms give the class $[{\Theta }^k]\in H^k(M,{\Omega }_M^k\otimes \mathrm{End}E)$. Even total degree of vector-bundle curvature and the noncommutative telescoping identity give ${\Theta }_1^k-{\Theta }_0^k=\bar{\partial }{\sum }_{j=0}^{k-1}{\Theta }_1^{j}a{\Theta }_0^{k-1-j}$ whenever ${\Theta }_1-{\Theta }_0=\bar{\partial }a$. Thus all powers are metric independent; no commutativity of endomorphisms is assumed.

For a nilpotent endomorphism $N$ of a vector space, the commutator action $\mathrm{ad}N:T\mapsto NT-TN$ on its endomorphisms is nilpotent. The commuting left and right multiplication maps give $(\mathrm{ad}N{)}^r(T)={\sum }_{j=0}^r(-1{)}^j\left(\genfrac{}{}{0ex}{}{r}{j}\right)N^{r-j}T{N}^j$. If $N^q=0$ and $r\ge 2q-1$, every summand vanishes. This works in every characteristic of a field and is useful when applying Engel theorem to an Adjoint representation.

For endomorphisms of a finite-dimensional complex vector space, suppose repeated commutation with $T$ annihilates $S$. Write the blocks of $S$ using the generalized eigenspaces of $T$. On $\mathrm{Hom}(V_{\lambda },V_{\mu })$, the operator $\mathrm{ad}T$ is $(\mu -\lambda )I$ plus a nilpotent endomorphism. If $\lambda \ne \mu$, it is invertible, so $(\mathrm{ad}T{)}^{N}S=0$ forces that off-diagonal block of $S$ to vanish. Thus $S$ preserves each generalized eigenspace. This allows a nilpotent Lie algebra representation to be split by generalized eigenvalues even when its acting operators do not commute.

A non-Hopfian group admits a surjective endomorphism which is not injective. The Baumslag-Solitar group $BS(2,3)$ has such an endomorphism $a\mapsto a^2$, $t\mapsto t$: its image contains $a^2$ and $a^3$, while the nonidentity commutator $[a,ta{t}^{-1}]$ lies in its kernel by Britton's lemma.

Two Riemannian metrics have the same Levi-Civita connection precisely when their relative positive self-adjoint endomorphism $A$ is parallel for that connection. Parallel transport preserves both metrics. Consequently agreement at one point implies agreement everywhere on a connected manifold. Without that agreement, constant scalar multiples always give examples; constant nonproportional Euclidean forms show that a common connection does not force scalar proportionality.