Proof about exact sequence 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$
 A: 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?
A: 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
