I think that is a fairly straightforward generalization. The point is that if you have a natural isomorphism F=>G (a natural monomorphism would actually suffice), then that natural isomorphism and G already determine F.
You should, if I'm not mistaken, obtain a functor from the indiscrete category whose underlying set is $\coprod_{f\colon\mathrm{Ob}(\mathcal{C})\rightarrow\mathrm{Ob}(\mathcal{C})}\prod_{x\in\mathcal{C}}\mathrm{Iso}(F(x),x)$ to the slice category $[\mathcal{C},\mathcal{C}]/1_{\mathcal{C}}$, which is an embedding onto the image, which is precisely the full subcategory on the …
You should, if I'm not mistaken, obtain a functor from the indiscrete category whose underlying set is $\coprod_{f\colon\mathrm{Ob}(\mathcal{C})\rightarrow\mathrm{Ob}(\mathcal{C})}\prod_{x\in\mathcal{C}}\mathrm{Iso}(F(x),x)$ to the slice category $[\mathcal{C},\mathcal{C}]/1_{\mathcal{C}}$, which is an embedding onto the image, which is precisely the full subcategory on the …
\begin{equation}
\limsup _{x\rightarrow x_{0} ;x\in E} f(x)=\inf\Bigl\{\sup \bigl\{f(x):x\in E\land |x-x_{0} |< \delta \bigr\} :\delta \in \mathbf{R}^{+}\Bigr\} \tag{A}
\end{equation}
We'll prove that the following are equivalent:
$\displaystyle \limsup _{E\ni x\rightarrow x_{0}} \ f( x) =L$
Fo...
Dirichlet's Theorem on arithmetic progressions is meant for the cosets $d\Bbb{Z} + a$ where $\gcd(a,d) = 1$.
However, what if you worked with fractional coefficients and thus "cosets" of the form $\dfrac{d}{c}\Bbb{Z} + \dfrac{a}{b}$, where the two fractions are given as reduced?
I think this woul...
Note: Please see this answer for more info on how this poses a security risk.
I'm sure you guys already know this, but allowing users to put images from any source in questions means that askers can get all the analytical data you can about their question.
So a malicious user could theoreticall...
