Lemma 1.3.11 of Categories & Sheaves. Having trouble proving $F_0$ is unique up to unique isomorphism. Lemma 1.3.11. Consider a functor $F: \mathcal{C} \to \mathcal{C'}$ and a full subcategory $\mathcal{C}_0'$ of $\mathcal{C}'$ such that for each $X \in \mathcal{C}$, there exists $Y \in \mathcal{C}_0'$ and an isomorphism $F(X) \simeq Y$.  Denote by $\iota'$ the embedding $\mathcal{C}_0' \to \mathcal{C}'$.  Then there exists a functor $F_0 : \mathcal{C} \to \mathcal{C}_0'$ and an isomorphism of functors $\theta_0: F \xrightarrow{\sim} \iota' \circ F_0$.  Moreover, $F_0$ is unique up to unique isomorphism.  More precisely, given another isomorphism $\theta_1 : F \xrightarrow{\sim}\iota' \circ F_1$, there exists a unique isomorphism of functors $\theta : F_1 \xrightarrow{\sim}F_0$ such that $\theta_0 = (\iota'\circ\theta)\circ \theta_1$.
Firstly, I'm not sure what they mean: do they mean $F_1$ has all the defining properties above that $F_0$ does?
I've proved everything up to the bolded part and not sure how to prove that.
Would this diagram be of any use in the proof?:

For example, what if we stuck two of these diagrams together "glued" at the $F$'s line, $F_1$'s on the topmost line.
Thanks.
 A: Firstly, I'm not sure what they mean: do they mean $F_1$ has all the defining properties above that $F_0$ does?
No, $F_1$ is an arbitrary functor $C\to C_0'$, such that there exists an isomorphism $\theta_1\colon F\to\iota'\circ F_1$.
I've proved everything up to the bolded part and not sure how to prove that.
Let $F_1$ be such a functor. Then the natural transformation $\theta_{1,0}=\theta_0\circ\theta_1^{-1}\colon\iota'\circ F_1\to\iota'\circ F_0$ is an isomorphism. Define the natural transformation $\theta\colon F_1\to F_0$ in the following way: $\theta(X)=\theta_{1,0}(X)$ for every $X\in C$. Note, that $\theta$ is an isomorphism because $C_0'$ is a full subcategory. Also $\theta_0=\theta_{1,0}\circ\theta_1=(\iota'\circ\theta)\circ\theta_1$. 
If $\theta'\colon F_1\to F_0$ is an another isomorphism with such property, then it is easy to see that $\iota'\circ\theta=\iota'\circ\theta'$, hence $\theta=\theta'$.

It may be easier to think about $\theta$ as the "corestriction" $\theta=\theta_{1,0}|^{C'_0}$.
Would this diagram be of any use in the proof?
This diagram shows that $F_0$ is a functor, so it is not useful in the proof of the statement in bold. I added the diagram which may appear helpful in visual understanding of this fact, but it is not a proof, of course.
A: 
Firstly, I'm not sure what they mean: do they mean $F_1$ has all the defining properties above that $F_0$ does?

Here is what we know $F_0$ satisfies: there exists a $\theta : F \overset{\sim}{\to} \iota' \circ F_0$.
Note that since composition by functors preserves isomorphisms of functors, this property is already invariant on isomorphism classes of functors.
The functors $F_1$ satisfying this property are unique up to isomorphism if knowing there exists $\theta_1 : F \overset{\sim}{\to} \iota' F_1$ implies there exists an $\eta : F_1 \overset{\sim}{\to} F_0$ down in $\mathcal{C}'_0$ such that the triangle
$$\require{AMScd}
\begin{CD}
F @>\theta_0>> \iota' F_0 \\
  @| @A \iota' \circ \eta AA\\
F @>\theta_1>> F_1
\end{CD}$$
commutes. 
They're unique up to unique isomorphism if this isomorphism $\eta : F' \simeq F_0$ is unique (while still satisfying the triangle above).
To see this, construct the isomorphism in $\mathcal{C}'$. Pointwise, it's given by inverting $\theta_0$ and then composing by $\theta_1$. Because $\iota'$ is full and faithful, you can lift this constructed isomorphism down to $\mathcal{C}_0'$. And it's unique because it was constructed uniquely from the very data you had to satisfy.
