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 Mar11 comment linear map combine conditions In (L4), set $b=1$ to get (L1), and $x_2=0$ to get (L2). Similar methods work for (L3) and (L5). Dec16 awarded Caucus Dec4 answered Functorial cofibrant replacement does not have to be fibration? Oct30 accepted Where to put the “such that”, given multiple quantifier Oct29 comment Where to put the “such that”, given multiple quantifier Well, it's as unnecessary as using it as a literal translation of "such that" ;) And since it's not part of the official syntax, it all boils down to personal opinion... Anyway, I think you gave a sufficient answer which I might very well accept in the near future. Oct29 comment Where to put the “such that”, given multiple quantifier Yeah, I get that point of view. Although, after finally figuring out what prenexified means, I think I agree more with GitGuts point of view. I.e. seeing the $:$ as a seperator of the string of quantifiers and the quantifier free part in a prenex normal form, instead of a literal "such that". Oct29 comment Where to put the “such that”, given multiple quantifier Yeah, i noticed the absence of $:$ in literature about logic, although the notation seems to be heavily used in papers and lecture notes in different fields. So it seems there's no real right and wrong here... Oct29 asked Where to put the “such that”, given multiple quantifier Oct20 comment bijective homomorphisms between non isomorphic posets, example , explanation needed @jhegedus There is no category theoretic isomorphism betwwen $P$ and $Q$. But there is one (non-trivial) from $P$ to $P$, mapping $b$ to $c$ and vice versa. Oct7 answered Identifyng objects in a category Oct7 comment Trying to understand significance of monoid as a one object category Considering your update: Define a category $\mathcal{N}$ via: $\operatorname{Ob}(\mathcal{N}) := \{*\}$, $\operatorname{Hom}_\mathcal{N}(*,*) := \{f_i|i\in\mathbb{N}\}$, and composition $f_i\circ f_j = f_{i+j}$. This might seem cheap, but it's easy to see that this $\mathcal{N}$ satisfies every axiom for a category (of course this is only one representant for the categorical monoid, since objects in $\mathcal{Cat}$ are only known up to isomorphisms). Oct6 comment Understanding the significance of a functor being full/faithful, especially with adjoints you can use $\operatorname{Hom}(x,y)\cong\operatorname{Hom}(Rx,Ry)\cong\operatorname{Hom}(LR‌​x,y)$ where the first iso comes from $R$ being fully faithful, and the second from R being right adjoint to $L$. Sep30 comment A question regarding Yoneda's lemma. I'm curious about your example though, seems like you misunderstood some part from the exercise. Sep29 comment Prove that the two polynomials intersect each other only at a single point Ok, i see. I parsed that wrong. Sep29 comment Prove that the two polynomials intersect each other only at a single point "What remains to show is that $D^K_1(\theta)−D^K_2(\theta)$ is neither positive nor negative for all $0<\theta<1/2$ given a $K$" Wait, what? calling the difference $f(\theta)$, and $\theta_0$ the point of intersection, then according to your graphs, $f(\theta)>0$ for $0<\theta<\theta_0$, and $f(\theta)<0$ for $1/2>\theta>\theta_0$ Sep18 comment 2-category in HoTT: chapter 9 from the HoTT book Questions should be self-contained, so it'd be nice if you'd actually post the exercise you're refering to. Sep10 awarded Curious Sep4 comment Proving a group, $G$, is a group action onto some set, $X$ composition has to hold for arbtrary elements $g_1$, $g_2\in A$. You have only checked the case $g_1 = g_2$ ($=g$, in your proof) Sep3 comment Right-adjoint to the inverse image functor @Patrick Way better than leaving the question unanswered. If you know the answer, you should consider writing one yourself Sep2 suggested rejected edit on Let $\displaystyle D= \{z: |x|<1\}$ which of the following is correct