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Oct
24
comment Group objects in category of $\mathcal{Set}$ are groups - How to prove it?
As pointed out in the answer, the points of a set $G$ can be naturally identified with the maps $X\to G$ where $X$ is any one-point set.
Oct
24
comment Group objects in category of $\mathcal{Set}$ are groups - How to prove it?
Here "defines a group" means that the set $G(Z)$ is a group under the binary operation $g\times g'=\mu\circ(g,g')$. (The notation is a bit confusing here; the first $(g,g')$ is a pair of morphisms that you're multiplying to define the group structure on $G(Z)$, whereas the $(g,g')$ in $\mu\circ(g,g')$ is the pair considered as a single map $Z\to G\times G$, by $z\mapsto(g(z),g'(z))$).
Oct
24
revised Quick question: Chern classes of Sym, Wedge, Hom, and Tensor
added 33 characters in body
Oct
24
comment For which $x$ is $e^x$ rational? Transcendental?
@pbs Thanks, that's clearer.
Oct
24
comment For which $x$ is $e^x$ rational? Transcendental?
All transcendental numbers are irrational, so I'm not sure what the role of the "or" is here. Did you mean to ask if $e^x$ is transcendental whenever $x\ne\log{a}$ for some $a\in\mathbb{Q}$?
Oct
24
comment Finding the co-ordinate vector
You seem to have swapped the roles of $x_1,x_2,x_3,x_4$ and $d,c,b,a$ from the question, which is maybe a little confusing!
Oct
24
comment Finding the co-ordinate vector
I think you might have a typo (or a confusion) in the last line as well - the solution is finding $x_1,x_2,x_3,x_4$, not $a,b,c,d$, which are arbitrary real numbers.
Oct
24
answered Finding the co-ordinate vector
Oct
24
comment Finding the co-ordinate vector
I don't think you meant for $v$ to be what you wrote - perhaps $v=a+bt+ct^2+dt^3$?
Oct
24
revised Finding the co-ordinate vector
deleted 18 characters in body
Oct
24
revised How to Prove that these Spaces are not Homotopically Equivalent
added 37 characters in body
Oct
24
comment Show that $G$ is $2$-connected but not necessarily Hamiltonian
Yes - this is fine! A graph is $n$-connected if you can delete any $k<n$ vertices without disconnecting it, so any $n$-connected graph is also $m$-connected for $m\leq n$. The Petersen graph is $3$-connected, hence $2$-connected.
Oct
24
comment Show that $G$ is $2$-connected but not necessarily Hamiltonian
Why do you think the Petersen graph is not $2$-connected?
Oct
24
revised Show that $G$ is $2$-connected but not necessarily Hamiltonian
added 8 characters in body; edited title
Oct
24
comment Are there definition of percent?
I don't think the problem here is whether or not $1\%=1/100$, but rather what "$5+4\%$" means. Does it mean $5$ whole units plus $4\%$ of an unit, so $5.04$ as in 5xum's answer, or $5$ plus $4\%$ of $5$, which is $5.2$? In the second case, it could more accurately be written as $5\times104\%$. In practice, you will probably have to work out from context (or ask somebody) what is meant.
Oct
23
comment Is continuity in topology well-defined?
@MathewGeorge The second identity might also fail; take the same map as in Hayden's example, then $f^{-1}[f[{x}]]=f^{-1}[{y_0}]=X$ for any point $x\in X$.
Oct
23
answered Applications of Baire's Threom
Oct
23
revised Transpose of composition of functions
added 91 characters in body
Oct
23
comment Transpose of composition of functions
Yes - as long as you can prove equality at some level, it's fine. I was conservative in the suggestion and went all the way down to evaluation on $U$, but as you point out, just playing with composition rules lets you prove equality after evaluation on $W^*$.
Oct
23
answered Transpose of composition of functions