Show that $T_{i-1}\circ d_i =d_i^\prime\circ T_i$ for $\{T_i\}$, a Chain Map from $C$ to $C'$. $\newcommand{\Hom}{\text{Hom}}$
The following exercise is from Brown's book, A Second Course in Linear Algebra.

Let $C=\{(V_i,d_i)\ |\ i\in\mathbb{Z}\}$ and $C'=\{(V_i^\prime,d_i^\prime)\ |\ i\in\mathbb{Z}\}$ are two chain complexes with $C'$ exact. Show that there exists $T\in\Hom(V_i,V_i^\prime)$ such that $T_{i-1}\circ d_i =d_i^\prime\circ T_i$ for all $i=1,2,\dots$ . The collection of linear maps $\{T_i\}$ is called a Chain Map from $C$ to $C'$.

Here, $\{V_i\}$ & $\{V_i^\prime\}$ are sequences of vector spaces and $\{d_i\}$ & $\{d_i^\prime\}$ are sequences of linear transformations, $d_i\in\Hom(V_i,V_{i-1})$ & $d_i^\prime\in\Hom(V_i^\prime,V_{i-1}^\prime)$.
I really have no idea how to prove the desired equality. I know that for exact chain complexes, we have $$\operatorname{Im}d_{i+1}^\prime = \ker d_i^\prime\ .$$ I am also aware that by definition, for any chain complex, $$d_{i+1}\circ d_i =0= d_{i+1}^\prime\circ d_i^\prime\ .$$
My plan was to use these facts to show that $T_{i-1}\circ d_i -d_i^\prime\circ T_i$ is the zero transformation for all integer $i$ but without success. I also tried rewriting as follows without further success:
\begin{equation}
(T_{i-1}\circ d_i -d_i^\prime\circ T_i)(V_i) = (T_{i-1}\circ d_i +d_{i+1}\circ d_i +d_{i+1}^\prime\circ d_i^\prime -d_i^\prime\circ T_i)(V_i).
\end{equation}
I have also thought about breaking $C'$ down to a collection of short exact sequences but I'm unsure how such a method is of any help.
Any hints or alternate approaches would be appreciated. TIA.
N.B.: From my Wikipedia and Internet search, I am aware that this is a well-known result in Homological Algebra. However, I haven't found any online article that specifically talks about chain maps and chain complexes of Vector Spaces. So, please don't mark this as a duplicate.
 A: As written, the exercise is trivial (just define $T_i=0$ for all $i$). And, many of the times, this is the only chain map.
Actually, the goal of the exercise is to construct a chain map from a given linear map. More precisely:
Exercise: Start with two chain complexes of vector spaces, $$
\require{AMScd}
\begin{CD}
\cdots @>>> V_2 @>{d_2}>> V_1 @>{d_1}>> V_0 @>>> 0
\end{CD}
$$ and $$
\require{AMScd}
\begin{CD}
\cdots @>>> W_2 @>{\partial_2}>> W_1 @>{\partial_1}>> W_0 @>>> 0,
\end{CD}
$$ where the last one is exact. If $T_0 \colon V_0 \to W_0$ is a linear map, show that there exists a family $T_1,T_2,\dots$ of linear maps that makes the following diagram commute. $$
\require{AMScd}
\begin{CD}
\cdots @>>> V_2 @>{d_2}>> V_1 @>{d_1}>> V_0 @>>> 0 \\
@. @VV{T_2}V @VV{T_1}V @VV{T_0}V \\
\cdots @>>> W_2 @>>{\partial_2}> W_1 @>>{\partial_1}> W_0 @>>> 0
\end{CD}
$$
For that, consider the following “elementary” result.
Lemma: Suppose $f \colon V \to X$ and $g \colon W \to X$ are linear maps. If $\operatorname{im} f \subseteq \operatorname{im} g$, then there exists a linear map $h \colon V \to W$ such that $gh = f$.
Proof: Consider a basis $\{e_i\}_{i \in I}$ for $V$. The condition $\operatorname{im} f \subseteq \operatorname{im} g$ implies that for each $i \in I$ there exists $w_i \in W$ with $f(e_i)=g(w_i)$. Thus, if $h \colon V \to W$ is the unique linear map such that $h(e_i)=w_i$ for all $i \in I$, then clearly $gh=f$. $\square$
Now, for the main exercise:
As $\operatorname{im}(T_0d_1) \subseteq W_0 = \operatorname{im} \partial_1$, by the lemma above there exists a linear map $T_1 \colon V_1 \to W_1$ such that $\partial_1 T_1 = T_0d_1$.
Next, as $\partial_1T_1d_2 = T_0d_1d_2 = T_00 = 0$, we have that $\operatorname{im}(T_1d_2) \subseteq \ker \partial_1 = \operatorname{im} \partial_2$, and so there exists a linear map $T_2 \colon V_2 \to W_2$ such that $\partial_2T_2 = T_1d_2$.
And so on.
