Power Objects $PX$ are Injective

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 

Comments

Post a Comment

Popular posts from this blog

90% visual proof of Contravariant Yoneda Lemma

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.
Read more

Elementary counting formula for the number of $2k$-separated prime pair averages in an interval.

Proof of 2k-separated prime pairs in an interval, counting formula Let all intervals be of integers. Fix a natural number k ≥ 1 k \geq 1 k ≥ 1 . Primorial p n # p_n\# p n ​ # is defined as usual to be p 1 p 2 ⋯ p n p_1 p_2 \cdots p_n p 1 ​ p 2 ​ ⋯ p n ​ . Let an undeclared p p p or q q q just mean any prime number. Lemma 1. The number of x ∈ [ 0 , y ] x \in [0,y] x ∈ [ 0 , y ] such that x 2 = k 2 ( m o d p i ) x^2 = k^2 \pmod {p_i} x 2 = k 2 ( mod p i ​ ) for no i = 1.. n i = 1..n i = 1.. n is given by: f ( y ) = ∑ d ∣ p n # ,   d ≤ y ( − 1 ) ω ( d ) ∑ c ∣ d ,   gcd ⁡ ( c , 2 k ) = 1 ⌊ y − ( z c , d ) ( d ) d ⌋ (0) f(y) = \sum_{d \mid p_n\#, \ \\ d \leq y } (-1)^{\omega(d)}\sum_{c \mid d, \ \\ \gcd(c,2k) = 1} \lfloor \dfrac{y - (z_{c,d})_{(d)}}{d}\rfloor \tag{0} f ( y ) = d ∣ p n ​ # ,   d ≤ y ∑ ​ ( − 1 ) ω ( d ) c ∣ d ,   g c d ( c , 2 k ) = 1 ∑ ​ ⌊ d y − ( z c , d ​ ) ( d ) ​ ​ ⌋ ( 0 ) Where ( z c , d ) ( d ) (z_{c,d})_{(d)} ( z ...
Read more

Understanding Product Maps $f \times g : A\times B \to C\times D$

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...
Read more