The proof of (1) is left to the reader. \medskip\noindent Let us take the rules of (2) and (3) as the definition of a simplicial object, call it $\tilde U$. We will show that it is an incarnation of $i_{k!}U$. This will prove (2), (3) at the same time. We have to show that given a morphism $f : U \to \text{sk}_kV$ there exists a unique morphism $\tilde f : \tilde U \to V$ which recovers $f$ upon taking the $k$-skeleton. From (1) we see that $f$ corresponds with a morphism $f_k : A \to V_k$ which maps into the kernel of $d^k_i$ for all $i$. For any surjective $\alpha : [n] \to [k]$ we set $\tilde f_\alpha : A \to V_n$ equal to the composition $\tilde f_\alpha = V(\alpha) \circ f_k : A \to V_n$. We define $\tilde f_n : \tilde U_n \to V_n$ as the sum of the $\tilde f_\alpha$ over $\alpha : [n] \to [k]$ surjective. Such a collection of $\tilde f_\alpha$ defines a morphism of simplicial objects if and only if for any $\varphi : [m] \to [n]$ the diagram $$ \xymatrix{ \bigoplus_{\alpha : [n] \to [k]\text{ surjective}} A \ar[r]_-{\tilde f_n} \ar[d]_{(3)} & V_n \ar[d]^{V(\varphi)} \\ \bigoplus_{\alpha' : [m] \to [k]\text{ surjective}} A \ar[r]^-{\tilde f_m} & V_m } $$ is commutative. Choosing $\varphi = \alpha$ shows our choice of $\tilde f_\alpha$ is uniquely determined by $f_k$. The commutativity in general may be checked for each summand of the left upper corner separately. It is clear for the summands corresponding to $\alpha$ where $\alpha \circ \varphi$ is surjective, because those get mapped by $\text{id}_A$ to the summand with $\alpha' = \alpha \circ \varphi$, and we have $\tilde f_{\alpha'} = V(\alpha') \circ f_k = V(\alpha \circ \varphi) \circ f_k = V(\varphi) \circ \tilde f_\alpha$. For those where $\alpha \circ \varphi$ is not surjective, we have to show that $V(\varphi) \circ \tilde f_\alpha = 0$. By definition this is equal to $V(\varphi) \circ V(\alpha) \circ f_k = V(\alpha \circ \varphi) \circ f_k$. Since $\alpha \circ \varphi$ is not surjective we can write it as $\delta^k_i \circ \psi$, and we deduce that $V(\varphi) \circ V(\alpha) \circ f_k = V(\psi) \circ d^k_i \circ f_k = 0$ see above. \medskip\noindent Let us take the rules of (4) and (5) as the definition of a simplicial object, call it $\tilde U$. We will show that it is an incarnation of $\text{cosk}_k U$. This will prove (4), (5) at the same time. The argument is completely dual to the proof of (2), (3) above, but we give it anyway. We have to show that given a morphism $f : \text{sk}_kV \to U$ there exists a unique morphism $\tilde f : V \to \tilde U$ which recovers $f$ upon taking the $k$-skeleton. From (1) we see that $f$ corresponds with a morphism $f_k : V_k \to A$ which is zero on the image of $s^{k - 1}_i$ for all $i$. For any injective $\beta : [k] \to [n]$ we set $\tilde f_\beta : V_n \to A$ equal to the composition $\tilde f_\beta = f_k \circ V(\beta) : V_n \to A$. We define $\tilde f_n : V_n \to \tilde U_n$ as the sum of the $\tilde f_\beta$ over $\beta : [k] \to [n]$ injective. Such a collection of $\tilde f_\beta$ defines a morphism of simplicial objects if and only if for any $\varphi : [m] \to [n]$ the diagram $$ \xymatrix{ V_n \ar[d]_{V(\varphi)} \ar[r]_-{\tilde f_n} & \bigoplus_{\beta : [k] \to [n]\text{ injective}} A \ar[d]^{(5)} \\ V_m \ar[r]^-{\tilde f_m} & \bigoplus_{\beta' : [k] \to [m]\text{ injective}} A } $$ is commutative. Choosing $\varphi = \beta$ shows our choice of $\tilde f_\beta$ is uniquely determined by $f_k$. The commutativity in general may be checked for each summand of the right lower corner separately. It is clear for the summands corresponding to $\beta'$ where $\varphi \circ \beta'$ is injective, because these summands get mapped into by exactly the summand with $\beta = \varphi \circ \beta'$ and we have in that case $\tilde f_{\beta'} \circ V(\varphi) = f_k \circ V(\beta') \circ V(\varphi) = f_k \circ V(\beta) = \tilde f_\beta$. For those where $\varphi \circ \beta'$ is not injective, we have to show that $\tilde f_{\beta'} \circ V(\varphi) = 0$. By definition this is equal to $f_k \circ V(\beta') \circ V(\varphi) = f_k \circ V(\varphi \circ \beta')$. Since $\varphi \circ \beta'$ is not injective we can write it as $\psi \circ \sigma^{k - 1}_i$, and we deduce that $f_k \circ V(\beta') \circ V(\varphi) = f_k \circ s^{k - 1}_i \circ V(\psi) = 0$ see above. \medskip\noindent The composition $i_{k!}U \to \text{cosk}_kU$ is the unique map of simplicial objects which is the identity on $A = U_k = (i_{k!}U)_k = (\text{cosk}_kU)_k$. Hence it suffices to check that the proposed rule defines a morphism of simplicial objects. To see this we have to show that for any $\varphi : [m] \to [n]$ the diagram $$ \xymatrix{ \bigoplus_{\alpha : [n] \to [k]\text{ surjective}} A \ar[d]_{(3)} \ar[r]_{(6)} & \bigoplus_{\beta : [k] \to [n]\text{ injective}} A \ar[d]^{(5)} \\ \bigoplus_{\alpha' : [m] \to [k]\text{ surjective}} A \ar[r]^{(6)} & \bigoplus_{\beta' : [k] \to [m]\text{ injective}} A } $$ is commutative. Now we can think of this in terms of matrices filled with only $0$'s and $1$'s as follows: The matrix of (3) has a nonzero $(\alpha', \alpha)$ entry if and only if $\alpha' = \alpha \circ \varphi$. Likewise the matrix of (5) has a nonzero $(\beta', \beta)$ entry if and only if $\beta = \varphi \circ \beta'$. The upper matrix of (6) has a nonzero $(\alpha, \beta)$ entry if and only if $\alpha \circ \beta = \text{id}_{[k]}$. Similarly for the lower matrix of (6). The commutativity of the diagram then comes down to computing the $(\alpha, \beta')$ entry for both compositions and seeing they are equal. This comes down to the following equality $$ \# \left\{ \beta \mid \beta = \varphi \circ \beta' \text{ and } \alpha \circ \beta = \text{id}_{[k]} \right\} = \# \left\{ \alpha' \mid \alpha' = \alpha \circ \varphi \text{ and } \alpha' \circ \beta' = \text{id}_{[k]} \right\} $$ whose proof may safely be left to the reader. \medskip\noindent Finally, we prove (7). This follows directly from Lemmas \ref{lemma-injective-map-simplicial-abelian}, \ref{lemma-recover-cosk}, \ref{lemma-recovering-U-for-real} and \ref{lemma-imshriek-abelian}.