Defining $\text{Ker }f\xrightarrow{j}A,\text{Im }f\xrightarrow{k} B,p:\text{Ker }c\to \text{Ker }f,q:\text{Im }c\to \text{Im }f,$ using elements?

Question

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Suppose I have a linear transformation or homomorphism map $f:A\to B,$ where $A,B$ can be groups, rings or modules. If I have the following two sequences of maps:

$\text{Ker }f\xrightarrow{j}A\xrightarrow{\sharp}A/\text{Ker }f\quad (*)$

$\text{Im }f\xrightarrow{k} B\xrightarrow{}B/\text{Im }f\quad (**)$

and a commutative square, with $a,c,d$ are all either linear transformations/homomorphic maps with $A,A',B,B'$ are either groups, rings, vector spaces or modules:

$$\begin{array}{ccccccccc} A' & \xrightarrow{a} & A \\ c\big\downarrow & & \big\downarrow f \\ B' & \xrightarrow{d} & B \\ \end{array},\quad (***)$$

From the above commutative square, I have the following two linear transformations/homomorphic maps: $p:\text{Ker }c\to \text{Ker }f$ and $q:\text{Im }c\to \text{Im }f.$

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What I would like to know is how do I describe the argument and image in terms of elements for the maps:

$\text{Ker }f\xrightarrow{j}A,\quad (1)$

$\text{Im }f\xrightarrow{k} B,\quad (2)$

$p:\text{Ker }c\to \text{Ker }f,\quad (3)$ and

$q:\text{Im }c\to \text{Im }f.\quad (4)$

For $(1),$ if $a\in \text{Ker }f,$ then $f(a)=0.$ Do the map $j$ in terms of elements is defined as: $a\mapsto j(a):=f(a)\mapsto j(f(a)):=0\mapsto j(0)?$

for $(2),$ if $b\in \text{Im }f,$ then $\exists a \in A, f(a)=b.$ Do the map $k$ in terms of elements is defined as: $b\mapsto k(b):=f(a)\mapsto k(f(a))?$

for $(3),$ again, if $a'\in \text{Ker }c,$ then the map $p$ in terms of elements is defined as:

$a'\mapsto p(a'):=c(a')\mapsto p(c(a')):=0\mapsto p(0).$ But $p(0)\in \text{Ker }f.$ So $p(c(a'))=p(0)=0=f(p(0))=f(p(c(a')))?$

for $(4)$ if $b'\in \text{Im}c,$ then $b=c(a'), a'\in A'.$ The map $q$ in terms of elements is defined as: $b'\mapsto q(b'):=c(a')\mapsto q(c(a')).$ But since $q(c(a'))\in \text{Im }f,$ there exists an $a\in A$ so that $q(c(a'))=f(a).$ Is this correct?

Thank you in advance.

Answer 1

Note that some of the things you write (in the generality you try to state them in) don't work...

Anyway: if you have a commutative diagram of functions (whether morphisms of algebraic structures or just functions) $$\begin{array}{rcl} A' & \stackrel{a}{\longrightarrow} & A \\ {\scriptstyle c}\big\downarrow & & \big\downarrow {\scriptstyle f} \\ B' & \stackrel{d}{\longrightarrow} & B \\ \end{array}$$ then it is necessarily the case that $d(\mathrm{Im}(c))\subseteq \mathrm{Im}(f)$: because if $b'\in\mathrm{Im}(c)$, then there exists $r\in A'$ such that $c(r)=b'$. Then $d(b') = d(c(r)) = f(a(r))\in\mathrm{Im}(f)$.

So if we restrict $d$ to $\mathrm{Im}(c)$, we get a map whose image is inside $\mathrm{Im}(f)$. So $d$ induces a map $q\colon\mathrm{Im}(c)\to\mathrm{Im}(f)$.

In the case where we are working with groups, or modules, or rings, so that the notion of "kernel" makes sense, if $x\in\mathrm{ker}(c)$, then $a(x)\in\ker(f)$. Because we have that (using $0$ to represent the identity element in the case of groups): $$f(a(x)) = d(c(x)) = d(0) = 0.$$

So $a(\ker(c)) \subseteq \ker(f)$, and therefore $a$ induces a map $p\colon\ker(c)\to\ker(f)$ which sends $x\in\ker(c)$ to $a(x)$.

$j$ is the inclusion map $\ker(f)\hookrightarrow A$.

$k$ is the inclusion map $\mathrm{Im}(f)\hookrightarrow B$.

So $p\colon \ker c\to\ker f$ is the restriction of $a$ to $\ker(c)$. Alternatively, it is the composition of the inclusion map $\ker(c)\hookrightarrow A'$ with $a$.

And $q\colon\mathrm{Im}(c)\to\mathrm{Im}(f)$ is the restriction of $d$ to $\mathrm{Im}(c)$. Alternatively, it is the composition of the inclusion $\mathrm{Im}(c)\hookrightarrow B'$ with $d$.

All of these maps have obvious formulas, where necessary in terms of $a$ and $d$.

Note that you also mention maps $A\to A/\ker(f)$ and $B\to B/\mathrm{Im}(f)$. The first map makes sense in the case of abelian groups, modules, vector spaces, groups, and rings, because you can always take a quotient modulo the kernel of a homomorphism. But the map $B\to B/\mathrm{Im}(f)$ does not make sense in that generality: it makes sense for abelian groups, modules, and vector spaces (because you can take the quotient modulo any subgroup/submodule/subspace). But they do not necessarily make sense in the setting of groups or rings, as the image of a group under a group homomorphism $A\to B$ need not be a normal subgroup of $B$; and the image of a ring homomorphism $A\to B$ is not generally an ideal (and in the case of unital rings with unital morphisms, it is never an ideal unless $f$ is surjective...)