EnjoysMath

By the fact that $\alpha_X : \text{Hom}_C(X,X) \to AX$ we have that $\alpha_X(\text{id}_X) =: u \in AX$ and we're done with mapping any natural map $\alpha : \text{Hom}_C(\cdot, Y) \to A$ to an element $u \in AX$. The sides of the triangular prism in the bottom diagram need to commute for each $f:Y\to X, g : Z \to Y$ in $C$. For one, the triangular endcaps must commute because $A, \text{Hom}_C(\cdot, X)$ are both contravariant functors. Next, the three square sides commute by naturality. We must then have that any time $\alpha_Y(f) = A(f)\circ u$, the diagram commutes.

See this blog post which proves that $d^2 = 0$ or indeed $\operatorname{im} d_{i+1} \subset \ker d_i$. Our goal here is to prove the opposite inclusion. By a trick mentioned in Lang, we can freely choose any element $z \in S$ and define $h: E_{i} \to E_{i+1}$ by linearly extending $h(x_0, \dots, x_i) = (z, x_0, \dots, x_i)$. We then need to prove that $dh + hd = \text{id}$ which would then imply that if $x \in \ker d$, then $x = (dh + hd)(x) = dh(x)$ or that $x$ is in the image of $d$. First we must get the chain map indices correct instead of succinctly working with the plain "$d$" notation: is what we think is meant by "$dh + hd$". So we need to prove that $d_{i+1} h_i + h_{i-1} d_i = \text{id}$. So here we go: $$ (d_{i+1} h_i + h_{i-1} d_i)(x) = d_{i+1} h_i(x) + h_{i-1} d_i(x) = \\ \sum_{j=0}^{i+1}(-1)^j(z, x_0, \dots,\widehat{x_j},\dots, x_{i}) + h_{i-1}\left(\sum_{j=0}^{i}(-1)^j(x_0, \dots, \widehat{x_j}, \dots, x_{i})\right) = $$ Notice that t...

The problem statement is mostly quoting from Lang's Algebra, top of page 764. Let $S$ be a set. For $i = 0, 1, 2,\dots$ let $E_i$ be the free $\Bbb{Z}$-module generated by the $(i+1)$-tuples $(x_0, \dots, x_{i})$ with each $x_j \in S$. The tuples $x := (x_0, \dots, x_i)$ form a basis for $E_i$ over $\Bbb{Z}$. Define $d_{i+1} : E_{i+1} \to E_i$ on the tuples and the linear (or "homomorphic") extension to all of $E_i$ uniquely determines a $\Bbb{Z}$-module homomorphism: $$ d_{i+1}(x) = d_{i+1}(x_0, \dots, x_{i+1}) = \sum_{j = 0}^{i+1} (-1)^j (x^{\hat{j}} = (x_0, \dots, \widehat{x_j}, \dots x_{i+1})) $$ Our goal is to prove that $d^2 = 0$ or that $d_{i}\circ d_{i+1} = 0$ the zero map. We will use a trick of breaking up a summation into two summations over $k \lt j$ and $k \gt j$. In our notation, we can't have $k = j$ because $j$ was already removed from the original tuple $(x_0, \dots, \widehat{x_j}, \dots, x_{i+1})$. That is our notation doesn't re-index ...

One way to understand these maps is to think of them as tuples of set maps $f: A \to C, g : B \to D$.  So if you're having trouble despite, the following categorical explanation, just think of them that way - the way from which the categorical construction is derived. Suppose that the products $A\times B$ and $C\times D$ exist in our category.  This means that by definition: Or in English: $A \xleftarrow{p_1} A \times B \xrightarrow{p_2} B$ is a product diagram if and only if for every glued in diagram $A \xleftarrow{x_1} X \xrightarrow{x_2} B$, there exists a unique map (i.e. UMP property here) $u: X \to A\times B$ such that everything commutes, i.e. $p_i u = x_i$ for each $i=1,2$. Now, we simply glue in the arrows $f$ and $g$: as well as the product $C \times D$ and its  projection maps: Can you finish the derivation from here?  If not, then please continue. Now, instead of the UMP for $A\times B$, use the UMP of $C\times D$ to get that there exis...

Proposition Let $\mathcal{E}$ be a topos and $X \in \mathcal{E}$ an object. If $PX$ is a power object of $X$, then $PX$ is injective . Proof By definition of power object, we have $\text{Hom}_{\mathcal{E}}(X \times Y, \Omega) \simeq \text{Hom}_{\mathcal{E}}(X\times Z, \Omega)$ where the isomorphism is in $\textbf{Set}$ (homsets are <i>sets</i>!), for any $X,Y,Z \in \mathcal{E}$.   Take 

Proposition Consider the following commutative square in any category $C$. That is, three of the maps are monomorphisms, while the fourth is $\text{id}$. Then the maps form a pullback square. Proof Suppose you draw a cone from an object $Z$ int $X$ and $Z$ like so: The only map from $u: W \to X$ such that $\text{id}\circ u = q$ is $u = q$. And this choice makes everything commute because since by assumption $m'p = m'' q = m'mq$ (by commutativity) which implies $p = mq$ since $m'$ is a monomorphism. Now, by symmetry of a pullback diagram along its diagonal, we do not have to check the case when $m$ and $\text{id}$ are swapped. $\blacksquare$

Proposition  Let $\mathcal{E}$ be a topos.  Then its subobject classifier $\Omega$ is an  injective object . Proof: Let $f: X \to \Omega$ and $m: X \rightarrowtail Y$ be two maps in $\mathcal{E}$.  Then since every map $f:X \to \Omega$  is the characteristic map of some mono  in $\mathcal{E}$ we have the following diagram, where $S \in \mathcal{E}$ is some object, and $m_f$ is the monomorphism induced by $f$. Now, composing $m$ and $m_f$ we get another monomorphism $m'$. Now, draw the following diagram: We know that a square of three monos and $\text{id}$ is a pullback square, therefore the inner square on the left is a pullback, in the above diagram. The right inner square is also a pullback, because $\Omega$ is a subobject classifier and $\chi_{m'}$ is the characteristic function of the mono $m'$. By the pullback-pasting lemma, we have that the perimeter of the rectangle is a pullback square. This means: is a pullback squa...