Proof.
The proof of (1) is left to the reader.
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.
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.
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.
Finally, we prove (7). This follows directly from Lemmas 14.18.7, 14.19.4, 14.21.3 and 14.21.9.
\square
Lemma 14.22.4. Let \mathcal{A} be an abelian category. Let A be an object of \mathcal{A} and let k be an integer \geq 0. Consider the simplicial object E defined by the following rules
E_ n = \bigoplus _\alpha A, where the sum is over \alpha : [n] \to [k + 1] whose image is either [k] or [k + 1].
Given \varphi : [m] \to [n] the map E_ n \to E_ m maps the summand corresponding to \alpha via \text{id}_ A to the summand corresponding to \alpha \circ \varphi , provided \mathop{\mathrm{Im}}(\alpha \circ \varphi ) is equal to [k] or [k + 1].
Then there exists a short exact sequence
0 \to K(A, k) \to E \to K(A, k + 1) \to 0
which is term by term split exact.
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