Proof about exact sequence

Question

Here is the sequence

I have the next sequence where the row is exact and $ h \circ f =0$. I need to prove that exists a homomorphism $k: C \to D$ uniquely determined such that $ k \circ g = h$

Answer 1

Hint:

Consider an element $x\in C$, lift it to a $\,\xi\in B$, and show that $h(\xi)$ does not depend on the lifting.

Can you take it from there?

Answer 2

Rather than try and write out the $\LaTeX$ for the diagram presented in the figure, I shall resort to a more "wordy" description. We have an exact sequence

$A \overset{f}{\longrightarrow} B \overset{g}{\longrightarrow} C \longrightarrow 0; \tag 1$

we are also given an object $D$ and a homomorphism

$h:B \longrightarrow D, \tag 2$

such that

$\text{im} f \subset \ker h,, \tag 3$

that is,

$h \circ f = 0. \tag 4$

Now to construct

$k:C \to D \tag 5$

such that

$h = k \circ g, \tag 6$

we first observe that the exactness of (1) implies that $g$ is surjective; thus for any

$c \in C \tag 7$

we have some

$b \in B \tag 8$

such that

$c = g(b); \tag 9$

we define $k(c)$ via

$k(c) = h(b), \tag{10}$

for then

$h(b) = k(g(b)). \tag{11}$

To ensure that $k$ is properly defined, we need to address the case of

$b_1, b_2 \in B, \tag{12}$

with

$g(b_1) = g(b_2) = c \in C; \tag{13}$

in such an event we must affirm that

$h(b_1) = h(b_2) \tag{14}$

in order that $k$ be well defined, i.e. that it depends only on $c$; if (13) holds, then

$g(b_1 - b_2) = 0, \tag{15}$

so

$b_1 - b_2 \in \ker g = \text{im} f; \tag{16}$

hence we have

$a \in f \tag{17}$

with

$b_1 - b_2 = f(a); \tag{18}$

thus,

$h(b_1) - h(b_2) = h(b_1 - b_2) = h \circ f(a) = 0 \tag{19}$

by virtue of (4); thus (14) is proved and hence $k$ is well-defined.

$k$ is clearly unique since by the above there is precisely one value of $h(b)$ with $k(c) = h(b)$.

Proof about exact sequence