hcju/stacksqa huggingface dataset:
For a morphism $\alpha : [n] \to [k + 1]$ we define $\alpha' : [n + 1] \to [k + 1]$ to be the map such that $\alpha'|_{[n]} = \alpha$ and $\alpha'(n + 1) = k + 1$. Note that if the image of $\alpha$ is $[k]$ or $[k + 1]$, then the image of $\alpha'$ is $[k + 1]$. Consider the family of maps $h_n : E_n \to E_{n + 1}$ which maps the summand corresponding to $\alpha$ to the summand corresponding to $\alpha'$ via the identity on $A$. Let us compute $d_{n + 1} \circ h_n - h_{n - 1} \circ d_n$. We will first do this in case the category $\mathcal{A}$ is the category of abelian groups. Let us use the notation $x_\alpha$ to indicate the element $x \in A$ in the summand of $E_n$ corresponding to the map $\alpha$ occurring in the index set. Let us also adopt the convention that $x_\alpha$ designates the zero element of $E_n$ whenever $\Im(\alpha)$ is not $[k]$ or $[k + 1]$. With these conventions we see that $$ d_{n + 1}(h_n(x_\alpha)) = \sum\nolimits_{i = 0}^{n + 1} (-1)^i x_{\alpha' \circ \delta^{n + 1}_i} $$ and $$ h_{n - 1}(d_n(x_\alpha)) = \sum\nolimits_{i = 0}^n (-1)^i x_{(\alpha \circ \delta_i^n)'} $$ It is easy to see that $\alpha' \circ \delta^{n + 1}_i = (\alpha \circ \delta_i^n)'$ for $i = 0, \ldots, n$. It is also easy to see that $\alpha' \circ \delta^{n + 1}_{n + 1} = \alpha$. Thus we see that $$ (d_{n + 1} \circ h_n - h_{n - 1} \circ d_n)(x_\alpha) = (-1)^{n + 1} x_\alpha $$ These identities continue to hold if $\mathcal{A}$ is any abelian category because they hold in the simplicial abelian group $[n] \mapsto \Hom(A, E_n)$; details left to the reader. We conclude that the identity map on $E$ is homotopic to zero, with homotopy given by the system of maps $h'_n = (-1)^{n + 1}h_n : E_n \to E_{n + 1}$. Hence we see that $E$ is acyclic, for example by Homology, Lemma \ref{homology-lemma-map-homology-homotopy}.
First, let us prove this if $k = 0$. In this case we have $K(A, 0)_n = A$ for all $n$. Furthermore, all the maps in this simplicial abelian group are $\text{id}_A$, in other words $K(A, 0)$ is the constant simplicial object with value $A$. The boundary maps $d_n = \sum_{i = 0}^n (-1)^i \text{id}_A = 0$ if $n$ odd and $ = \text{id}_A$ if $n$ is even. Thus $s(K(A, 0))$ looks like this $$ \ldots \to A \xrightarrow{0} A \xrightarrow{1} A \xrightarrow{0} A \to 0 $$ and the result is clear. \medskip\noindent Next, we prove the result for all $k$ by induction. Given the result for $k$ consider the short exact sequence $$ 0 \to K(A, k) \to E \to K(A, k + 1) \to 0 $$ from Lemma \ref{lemma-extension}. By Lemma \ref{lemma-abelian} the associated sequence of chain complexes is exact. By Lemma \ref{lemma-homology-extension} we see that $s(E)$ is acyclic. Hence the result for $k + 1$ follows from the long exact sequence of homology, see Homology, Lemma \ref{homology-lemma-long-exact-sequence-chain}.
It is clear that $N(K(A, k))_i = 0$ when $i < k$ because $K(A, k)_i = 0$ in that case. It is clear that $N(K(A, k))_k = A$ since $K(A, k)_{k - 1} = 0$ and $K(A, k)_k = A$. For $i > k$ we have $N(K(A, k))_i = 0$ by Lemma \ref{lemma-imshriek-abelian} and the definition of $K(A, k)$, see Definition \ref{definition-eilenberg-maclane}.
Define $D(U)_n$ to be the image of $$ \bigoplus\nolimits_{\varphi : [n] \to [m]\text{ surjective}, \ m < n} N(U_m) \xrightarrow{\bigoplus U(\varphi)} U_n $$ which is a subobject of $U_n$ complementary to $N(U_n)$ according to Lemma \ref{lemma-splitting-abelian-category} and Definition \ref{definition-split}. We show that $D(U)$ is a subcomplex. Pick a surjective map $\varphi : [n] \to [m]$ with $m < n$ and consider the composition $$ N(U_m) \xrightarrow{U(\varphi)} U_n \xrightarrow{d_n} U_{n - 1}. $$ This composition is the sum of the maps $$ N(U_m) \xrightarrow{U(\varphi \circ \delta^n_i)} U_{n - 1} $$ with sign $(-1)^i$, $i = 0, \ldots, n$. \medskip\noindent First we will prove by ascending induction on $m$, $0 \leq m < n - 1$ that all the maps $U(\varphi \circ \delta^n_i)$ map $N(U_m)$ into $D(U)_{n - 1}$. (The case $m = n - 1$ is treated below.) Whenever the map $\varphi \circ \delta^n_i : [n - 1] \to [m]$ is surjective then the image of $N(U_m)$ under $U(\varphi \circ \delta^n_i)$ is contained in $D(U)_{n - 1}$ by definition. If $\varphi \circ \delta^n_i : [n - 1] \to [m]$ is not surjective, set $j = \varphi(i)$ and observe that $i$ is the unique index whose image under $\varphi$ is $j$. We may write $\varphi \circ \delta^n_i = \delta^m_j \circ \psi \circ \delta^n_i$ for some $\psi : [n - 1] \to [m - 1]$. Hence $U(\varphi \circ \delta^n_i) = U(\psi \circ \delta^n_i) \circ d^m_j$ which is zero on $N(U_m)$ unless $j = m$. If $j = m$, then $d^m_m(N(U_m)) \subset N(U_{m - 1})$ and hence $U(\varphi \circ \delta^n_i)(N(U_m)) \subset U(\psi \circ \delta^n_i)(N(U_{m - 1}))$ and we win by induction hypothesis. \medskip\noindent To finish proving that $D(U)$ is a subcomplex we still have to deal with the composition $$ N(U_m) \xrightarrow{U(\varphi)} U_n \xrightarrow{d_n} U_{n - 1}. $$ in case $m =  n - 1$. In this case $\varphi = \sigma^{n - 1}_j$ for some $0 \leq j \leq n - 1$ and $U(\varphi) = s^{n - 1}_j$. Thus the composition is given by the sum $$ \sum (-1)^i d^n_i \circ s^{n - 1}_j $$ Recall from Remark \ref{remark-relations} that $d^n_j \circ s^{n - 1}_j = d^n_{j + 1} \circ s^{n - 1}_j = \text{id}$ and these drop out because the corresponding terms have opposite signs. The map $d^n_n \circ s^{n - 1}_j$, if $j < n - 1$, is equal to $s^{n - 2}_j \circ d^{n - 1}_{n - 1}$. Since $d^{n - 1}_{n - 1}$ maps $N(U_{n - 1})$ into $N(U_{n - 2})$, we see that the image $d^n_n ( s^{n - 1}_j (N(U_{n - 1}))$ is contained in $s^{n - 2}_j(N(U_{n - 2}))$ which is contained in $D(U_{n - 1})$ by definition. For all other combinations of $(i, j)$ we have either $d^n_i \circ s^{n - 1}_j = s^{n - 2}_{j - 1} \circ d^{n - 1}_i$ (if $i < j$), or $d^n_i \circ s^{n - 1}_j = s^{n - 2}_j \circ d^{n - 1}_{i - 1}$ (if $n > i > j + 1$) and in these cases the map is zero because of the definition of $N(U_{n - 1})$.
Note that the result holds for $K(A, k)$ for any object $A$ and any $k \geq 0$, by Lemmas \ref{lemma-homology-eilenberg-maclane} and \ref{lemma-N-K}. Consider the hypothesis $IH_{n, m}$: for all $V$ such that $V_j = 0$ for $j \leq m$ and all $i \leq n$ the map $N(V) \to s(V)$ induces an isomorphism $H_i(N(V)) \to H_i(s(V))$. \medskip\noindent To start of the induction, note that $IH_{n, n}$ is trivially true, because in that case $N(V)_n = 0$ and $s(V)_n = 0$. \medskip\noindent Assume $IH_{n, m}$, with $m \leq n$. Pick a simplicial object $V$ such that $V_j = 0$ for $j < m$. By Lemma \ref{lemma-eilenberg-maclane-object} and Definition \ref{definition-eilenberg-maclane} we have $K(V_m, m) = i_{m!} \text{sk}_mV$. By Lemma \ref{lemma-n-skeleton-abelian} the natural morphism $$ K(V_m, m) = i_{m!} \text{sk}_mV \to V $$ is injective. Thus we get a short exact sequence $$ 0 \to K(V_m, m) \to V \to W \to 0 $$ for some $W$ with $W_i = 0$ for $i = 0, \ldots, m$. This short exact sequence induces a morphism of short exact sequence of associated complexes $$ \xymatrix{ 0 \ar[r] & N(K(V_m, m)) \ar[r] \ar[d] & N(V) \ar[r] \ar[d] & N(W) \ar[r] \ar[d] & 0 \\ 0 \ar[r] & s(K(V_m, m)) \ar[r] & s(V) \ar[r] & s(W) \ar[r] & 0 } $$ see Lemmas \ref{lemma-s-exact} and \ref{lemma-N-exact}. Hence we deduce the result for $V$ from the result on the ends.
The first assertion is trivial. For the second, consider the map of simplicial sets $\Delta[1] \times \Delta[1] \longrightarrow \Delta[1]$ which in degree $n$ assigns to a pair $(\beta_1, \beta_2)$, $\beta_i : [n] \to [1]$ the morphism $\beta : [n] \to [1]$ defined by the rule $$ \beta(i) = \max\{\beta_1(i), \beta_2(i)\}. $$ It is a morphism of simplicial sets, because the action $\Delta[1](\varphi) : \Delta[1]_n \to \Delta[1]_m$ of $\varphi : [m] \to [n]$ is by precomposing. Clearly, using notation from Section \ref{section-homotopy}, we have $\beta = \beta_1$ if $\beta_2 = \alpha^n_0$ and $\beta = \alpha^n_{n + 1}$ if $\beta_2 = \alpha^n_{n + 1}$. This implies easily that the induced morphism $$ U \times \Delta[1] \times \Delta[1] \longrightarrow U \times \Delta[1] $$ of Lemma \ref{lemma-back-to-U} is a homotopy from $\text{id}_{U \times \Delta[1]}$ to $e_0 \circ \pi$. Similarly for $e_1 \circ \pi$ (use minimum instead of maximum).
Set $g^0 = \text{id}_Y$ and $g^1 = s \circ f$. We use the morphisms \begin{eqnarray*} Y \times_X \ldots \times_X Y \times \Mor([n], [1]) & \to & Y \times_X \ldots \times_X Y \\ (y_0, \ldots, y_n) \times \alpha & \mapsto & (g^{\alpha(0)}(y_0), \ldots, g^{\alpha(n)}(y_n)) \end{eqnarray*} where we use the functor of points point of view to define the maps. Another way to say this is to say that $h_{n, 0} = \text{id}$, $h_{n, n + 1} = (s \circ f)^{n + 1}$ and $h_{n, i} = \text{id}_Y^{i + 1} \times (s \circ f)^{n + 1 - i}$. We leave it to the reader to show that these satisfy the relations of Lemma \ref{lemma-relations-homotopy}. Hence they define the desired homotopy. See also Remark \ref{remark-homotopy-better} which shows that we do not need to assume anything else on the category $\mathcal{C}$.
If $h_t = (h_{t, n , i})$ is a homotopy from $a_t$ to $b_t$ (see Remark \ref{remark-homotopy-better}), then $h = (\prod_t h_{t, n, i})$ is a homotopy from $\prod a_t$ to $\prod b_t$. This proves (2). \medskip\noindent Proof of (3). Choose $t \in T$. There exists an integer $n \geq 0$ and a chain $a_t = a_{t, 0}, a_{t, 1}, \ldots, a_{t, n} = b_t$ such that for every $1 \leq i \leq n$ either there is a homotopy from $a_{t, i - 1}$ to $a_{t, i}$ or there is a homotopy from $a_{t, i}$ to $a_{t, i - 1}$. If $n = 0$, then we pick another $t$. (We're done if $a_t = b_t$ for all $t \in T$.) So assume $n > 0$. By Example \ref{example-trivial-homotopy} there are is a homotopy from $b_{t'}$ to $b_{t'}$ for all $t' \in T \setminus \{t\}$. Thus by (2) there is a homotopy from $a_{t, n - 1} \times \prod_{t'} b_{t'}$ to $b$ or there is a homotopy from $b$ to $a_{t, n - 1} \times \prod_{t'} b_{t'}$. In this way we can decrease $n$ by $1$. This proves (3). \medskip\noindent Part (1) follows from part (3) and the definitions.
We may choose a sequence $a = a_0, a_1, \ldots, a_n = b$ of morphisms from $U$ to $V$ such that for each $i = 1, \ldots, n$ either there is a homotopy from $a_i$ to $a_{i - 1}$ or there is a homotopy from $a_{i - 1}$ to $a_i$. The calculation above shows that in this case either $s(a_i)$ is homotopic to $s(a_{i - 1})$ as a map of chain complexes or $s(a_{i - 1})$ is homotopic to $s(a_i)$ as a map of chain complexes. Of course, these things are equivalent and moreover being homotopic is an equivalence relation on the set of maps of chain complexes, see Homology, Section \ref{homology-section-complexes}. This proves that $s(a)$ and $s(b)$ are homotopic as maps of chain complexes. \medskip\noindent Next, we turn to $N(a)$ and $N(b)$. It follows from Lemma \ref{lemma-decompose-associated-complexes} that $N(a)$, $N(b)$ are compositions $$ N(U) \to s(U) \to s(V) \to N(V) $$ where we use $s(a)$, $s(b)$ in the middle. Hence the assertion follows from Homology, Lemma \ref{homology-lemma-compose-homotopy}.
In case $\mathcal{C}$ has finite products, then $\mathcal{C}^{opp}$ has finite coproducts and we may use Definitions \ref{definition-homotopy-cosimplicial} and \ref{definition-homotopy} instead of Remarks \ref{remark-homotopy-cosimplicial-better} and \ref{remark-homotopy-better}. In this case $h : U \to \Hom(\Delta[1], V)$ is the same as a morphism $h' : \Hom(\Delta[1], V)' \to U'$. Since products and coproducts get switched too, it is immediate that $(\Hom(\Delta[1], V))' = V' \times \Delta[1]$. Moreover, the ``primed'' version of the morphisms $e_0, e_1 : \Hom(\Delta[1], V) \to V$ are the morphisms $e_0, e_1 : V' \to \Delta[1] \times V$. Thus $e_0 \circ h = a$ translates into $h' \circ e_0 = a'$ and similarly $e_1 \circ h = b$ translates into $h' \circ e_1 = b'$. This proves the lemma in this case. \medskip\noindent In the general case, one needs to translate the relations given by (\ref{equation-property-homotopy-cosimplicial}) into the relations given in Lemma \ref{lemma-relations-homotopy}. We omit the details. \medskip\noindent The final assertion is formal from the equivalence of (1) and (2).
Let $(-)' : \mathcal{A} \to \mathcal{A}^{opp}$ be the contravariant functor $A \mapsto A$. By Lemma \ref{lemma-push-outs-simplicial-object-w-section} the maps $a'$ and $b'$ are homotopic. By Lemma \ref{lemma-homotopy-s-N} we see that $s(a')$ and $s(b')$ are homotopic maps of chain complexes. Since $s(a') = (s(a))'$ and $s(b') = (s(b))'$ we conclude that also $s(a)$ and $s(b)$ are homotopic by applying the additive contravariant functor $(-)'' : \mathcal{A}^{opp} \to \mathcal{A}$. The result for the $Q$-complexes follows in the same manner using that $Q(U)' = N(U')$.
Denote $(p_n, q_n) : B_n \to A_n \oplus A_n$ the morphism $\pi_n$ of Homology, Lemma \ref{homology-lemma-ses-termwise-split}. Also write $(a, b) : A \oplus A \to B$, and $r : B \to C$ for the maps in the short exact sequence. Write the factorization of $\delta(s)$ as $\delta(s) = (1, -1) \circ f$. This means that $p_{n - 1} \circ d_{B, n} \circ s_n = f_n$, and $q_{n - 1} \circ d_{B, n} \circ s_n = - f_n$, and Set $B_n \to \diamond A_n = A_n \oplus A_n \oplus A_{n - 1}$ equal to $(p_n, q_n, f_n \circ r_n)$. \medskip\noindent Now we have to check that this actually defines a morphism of complexes. We will only do this in the case of abelian groups. Pick $x \in B_n$. Then $x = a_n(x_1) + b_n(x_2) + s_n(x_3)$ and it suffices to show that our definition commutes with differential for each term separately. For the term $a_n(x_1)$ we have $(p_n, q_n, f_n \circ r_n)(a_n(x_1)) = (x_1, 0, 0)$ and the result is obvious. Similarly for the term $b_n(x_2)$. For the term $s_n(x_3)$ we have \begin{eqnarray*} (p_n, q_n, f_n \circ r_n)(d_n(s_n(x_3))) & = & (p_n, q_n, f_n \circ r_n)( \\ & & \ \ \ \ \ a_n(f_n(x_3)) - b_n(f_n(x_3)) + s_n(d_n(x_3))) \\ & = & (f_n(x_3), -f_n(x_3), f_n(d_n(x_3))) \end{eqnarray*} by definition of $f_n$. And \begin{eqnarray*} d_n(p_n, q_n, f_n \circ r_n)(s_n(x_3)) & = & d_n(0, 0, f_n(x_3)) \\ & = & (f_n(x_3), - f_n(x_3), d_{A[-1], n}(f_n(x_3))) \end{eqnarray*} The result follows as $f$ is a morphism of complexes.
Suppose that $Z \not = W$. Let $n$ be the smallest integer such that $Z_n \not = W_n$. Let $x \in W_n$, $x \not \in Z_n$. Denote $Z' \subset W$ the simplicial subset containing $Z$, $x$, and all degeneracies of $x$. Let $\varphi : \Delta[n] \to Z'$ be the morphism corresponding to $x$ (Lemma \ref{lemma-simplex-map}). Then $\varphi|_{\partial \Delta[n]}$ maps into $Z$ as all the nondegenerate simplices of $\partial \Delta[n]$ end up in $Z$. By assumption we can extend $b \circ \varphi|_{\partial \Delta[n]}$ to $\beta : \Delta[n] \to X$. By Lemma \ref{lemma-glue-simplex} the simplicial set $Z'$ is the pushout of $\Delta[n]$ and $Z$ along $\partial \Delta[n]$. Hence $b$ and $\beta$ define a morphism $b' : Z' \to X$. In other words, we have extended the morphism $b$ to a bigger simplicial subset of $Z$. \medskip\noindent The proof is finished by an application of Zorn's lemma (omitted).
By Lemma \ref{lemma-trivial-kan} we can choose an right inverse $g : Y \to X$ to $f$. Consider the diagram $$ \xymatrix{ \partial \Delta[1] \times X \ar[d] \ar[r] & X \ar[d] \\ \Delta[1] \times X \ar[r] \ar@{-->}[ru] & Y } $$ Here the top horizontal arrow is given by $\text{id}_X$ and $g \circ f$ where we use that $(\partial \Delta[1] \times X)_n = X_n \amalg X_n$ for all $n \geq 0$. The bottom horizontal arrow is given by the map $\Delta[1] \to \Delta[0]$ and $f : X \to Y$. The diagram commutes as $f \circ g \circ f = f$. By Lemma \ref{lemma-trivial-kan} we can fill in the dotted arrow and we win.
Consider a commutative solid diagram $$ \xymatrix{ \Lambda_k[n] \ar[r]_a \ar[d] & X \ar[d] \\ \Delta[n] \ar[r]^b \ar@{-->}[ru] & Y } $$ as in Definition \ref{definition-kan}. The map $a$ corresponds to $x_0, \ldots, \hat x_k, \ldots, x_n \in X_{n - 1}$ satisfying $d_i x_j = d_{j - 1} x_i$ for $i < j$, $i, j \not = k$. The map $b$ corresponds to an element $y \in Y_n$ such that $d_iy = f(x_i)$ for $i \not = k$. Our task is to produce an $x \in X_n$ such that $d_ix = x_i$ for $i \not = k$ and $f(x) = y$. \medskip\noindent Since $f$ is termwise surjective we can find $x \in X_n$ with $f(x) = y$. Replace $y$ by $0 = y - f(x)$ and $x_i$ by $x_i - d_ix$ for $i \not = k$. Then we see that we may assume $y = 0$. In particular $f(x_i) = 0$. In other words, we can replace $X$ by $\Ker(f) \subset X$ and $Y$ by $0$. In this case the statement become Lemma \ref{lemma-simplicial-group-kan}.
Consider a commutative solid diagram $$ \xymatrix{ \partial \Delta[n] \ar[r]_a \ar[d] & X \ar[d] \\ \Delta[n] \ar[r]^b \ar@{-->}[ru] & Y } $$ as in Definition \ref{definition-trivial-kan}. The map $a$ corresponds to $x_0, \ldots, x_n \in X_{n - 1}$ satisfying $d_i x_j = d_{j - 1} x_i$ for $i < j$. The map $b$ corresponds to an element $y \in Y_n$ such that $d_iy = f(x_i)$. Our task is to produce an $x \in X_n$ such that $d_ix = x_i$ and $f(x) = y$. \medskip\noindent Since $f$ is termwise surjective we can find $x \in X_n$ with $f(x) = y$. Replace $y$ by $0 = y - f(x)$ and $x_i$ by $x_i - d_ix$. Then we see that we may assume $y = 0$. In particular $f(x_i) = 0$. In other words, we can replace $X$ by $\Ker(f) \subset X$ and $Y$ by $0$. This works, because by Homology, Lemma \ref{homology-lemma-long-exact-sequence-chain} the homology of the chain complex associated to $\Ker(f)$ is zero and hence $\Ker(f) \to 0$ induces a quasi-isomorphism on associated chain complexes. \medskip\noindent Since $X$ is a Kan complex (Lemma \ref{lemma-simplicial-group-kan}) we can find $x \in X_n$ with $d_i x = x_i$ for $i = 0, \ldots, n - 1$. After replacing $x_i$ by $x_i - d_ix$ for $i = 0, \ldots, n$ we may assume that $x_0 = x_1 = \ldots = x_{n - 1} = 0$. In this case we see that $d_i x_n = 0$ for $i = 0, \ldots, n - 1$. Thus $x_n \in N(X)_{n - 1}$ and lies in the kernel of the differential $N(X)_{n - 1} \to N(X)_{n - 2}$. Here $N(X)$ is the normalized chain complex associated to $X$, see Section \ref{section-complexes}. Since $N(X)$ is quasi-isomorphic to $s(X)$ (Lemma \ref{lemma-quasi-isomorphism}) and thus acyclic we find $x \in N(X_n)$ whose differential is $x_n$. This $x$ answers the question posed by the lemma and we are done.
In this proof we will write $H_n(Z) = H_n(s(Z)) = H_n(N(Z))$ when $Z$ is a simplicial abelian group, with $s$ and $N$ as in Section \ref{section-complexes}. Let $\mathbf{Z}[X]$ denote the free abelian group on $X$ viewed as a simplicial set and similarly for $\mathbf{Z}[Y]$. Consider the commutative diagram $$ \xymatrix{ \mathbf{Z}[X] \ar[r]_g \ar[d] & \mathbf{Z}[Y] \ar[d] \\ X \ar[r]^f & Y } $$ of simplicial abelian groups. Since taking the free abelian group on a set is a functor, we see that the horizontal arrow is a homotopy equivalence of simplicial abelian groups, see Lemma \ref{lemma-functorial-homotopy}. By Lemma \ref{lemma-homotopy-equivalence-s-N} we see that $H_n(g) : H_n(\mathbf{Z}[X]) \to H_n(\mathbf{Z}[Y])$ is bijective for all $n \geq 0$. \medskip\noindent Let $\xi \in H_n(Y)$. By definition of $N(Y)$ we can represent $\xi$ by an element $y \in N(Y_n)$ whose boundary is zero. This means $y \in Y_n$ with $d^n_0(y) = \ldots = d^n_{n - 1}(y) = 0$ because $y \in N(Y_n)$ and $d^n_n(y) = 0$ because the boundary of $y$ is zero. Denote $0_n \in Y_n$ the zero element. Then we see that $$ \tilde y = [y] - [0_n] \in (\mathbf{Z}[Y])_n $$ is an element with $d^n_0(\tilde y) = \ldots = d^n_{n - 1}(\tilde y) = 0$ and $d^n_n(\tilde y) = 0$. Thus $\tilde y$ is in $N(\mathbf{Z}[Y])_n$ has boundary $0$, i.e., $\tilde y$ determines a class $\tilde \xi \in H_n(\mathbf{Z}[Y])$ mapping to $\xi$. Because $H_n(\mathbf{Z}[X]) \to H_n(\mathbf{Z}[Y])$ is bijective we can lift $\tilde \xi$ to a class in $H_n(\mathbf{Z}[X])$. Looking at the commutative diagram above we see that $\xi$ is in the image of $H_n(X) \to H_n(Y)$. \medskip\noindent Let $\xi \in H_n(X)$ be an element mapping to zero in $H_n(Y)$. Exactly as in the previous parapgraph we can represent $\xi$ by an element $x \in N(X_n)$ whose boundary is zero, i.e., $d^n_0(x) = \ldots = d^n_{n - 1}(x) = d^n_n(x) = 0$. In particular, we see that $[x] - [0_n]$ is an element of $N(\mathbf{Z}[X])_n$ whose boundary is zero, whence defines a lift $\tilde \xi \in H_n(\mathbf{Z}[x])$ of $\xi$. The fact that $\xi$ maps to zero in $H_n(Y)$ means there exists a $y \in N(Y_{n + 1})$ whose boundary is $f_n(x)$. This means $d^{n + 1}_0(y) = \ldots = d^{n + 1}_n(y) = 0$ and $d^{n + 1}_{n + 1}(y) = f(x)$. However, this means exactly that $z = [y] - [0_{n + 1}]$ is in $N(\mathbf{Z}[y])_{n + 1}$ and $$ g([x] - [0_n]) = [f(x)] - [0_n] = \text{boundary of }z $$ This proves that $\tilde \xi$ maps to zero in $H_n(\mathbf{Z}[y])$. As $H_n(\mathbf{Z}[X]) \to H_n(\mathbf{Z}[Y])$ is bijective we conclude $\tilde \xi = 0$ and hence $\xi = 0$.
Consider a solid diagram $$ \xymatrix{ \partial \Delta[k] \ar[r] \ar[d] & V \ar[d] \\ \Delta[k] \ar[r] \ar@{-->}[ru] & U } $$ as in Definition \ref{definition-trivial-kan}. Let $x \in U_k$ be the $k$-simplex corresponding to the lower horizontal arrow. If $k \leq n$ then the dotted arrow is the one corresponding to a lift $y \in V_k$ of $x$; the diagram will commute as the other nondegenerate simplices of $\Delta[k]$ are in degrees $< k$ where $f$ is an isomorphism. If $k > n$, then by conditions (3) and (4) we have (using adjointness of skeleton and coskeleton functors) $$ \Mor(\Delta[k], U) = \Mor(\text{sk}_n\Delta[k], \text{sk}_nU) = \Mor(\text{sk}_n\partial\Delta[k], \text{sk}_nU) = \Mor(\partial \Delta[k], U) $$ and similarly for $V$ because $\text{sk}_n\Delta[k] = \text{sk}_n\partial\Delta[k]$ for $k > n$. Thus we obtain a unique dotted arrow fitting into the diagram in this case also.
[First proof] Let $W$ be the $n$-truncated simplicial set with $W_i = U_i$ for $i < n$ and $W_n = U_n / \sim$ where $\sim$ is the equivalence relation generated by $f^0(y) \sim f^1(y)$ for $y \in V_n$. This makes sense as the morphisms $U(\varphi) : U_n \to U_i$ corresponding to $\varphi : [i] \to [n]$ for $i < n$ factor through the quotient map $U_n \to W_n$ because $f^0$ and $f^1$ are morphisms of simplicial sets and equal in degrees $< n$. Next, we upgrade $W$ to a simplicial set by taking $\text{cosk}_n W$. By Lemma \ref{lemma-section} the morphism $g : U \to W$ is a trivial Kan fibration. Observe that $g \circ f^0 = g \circ f^1$ by construction and denote this morphism $f : V \to W$. Consider the diagram $$ \xymatrix{ \partial \Delta[1] \times V \ar[rr]_{f^0, f^1} \ar[d] & & U \ar[d] \\ \Delta[1] \times V \ar[rr]^f \ar@{-->}[rru] & & W } $$ By Lemma \ref{lemma-trivial-kan} the dotted arrow exists and the proof is done.
Observe that $U$ fits into a cartesian square $$ \xymatrix{ U \ar[d] \ar[r] & \text{cosk}_0(B) \ar[d] \\ B \ar[r] & \text{cosk}_0(A) } $$ Since the right vertical arrow is a trivial Kan fibration by Lemma \ref{lemma-section}, so is the left by Lemma \ref{lemma-trivial-kan-base-change}.