I want to know, Why the zero morphism is independent of the zero object. I have $0 \in \mathcal{C}$ the zero object and $0_{X,Y}: X \to 0 \to Y$ the zero morphism. I am very confused to prove that the zero morphism is independent of the zero object. Can anybody give me an idea?
122563
I want to prove that zero morphism does not depend on zero object. Let a zero object $0 \in \mathcal{C}$.  I supposed that exists another zero object $0' \in \mathcal{C}$, and I considered $0: A \to B$ and also $$A \to 0 \to 0' \to B$$ But I don't know to prove that zero morphism $0: A \to B$ not does not depend on zero object. Can anybody give me a suggestion?
22490
I was just wondering whether anyone knows an answer to the following:  Suppose that ${\mathbb F}$ is a subfield of a field ${\mathbb G}$ and that $v_1,\ldots ,v_k$ are linearly independent vectors in ${\mathbb F}^n$ (over $\mathbb F$). Is it necessarily true that $v_1,\ldots ,v_k$ are also linearly independent when considered as vectors in ${\mathbb G}^n$ (over $\mathbb G$})?  Any help would be much appreciated. Thanks!
80542
If $\varphi:R^{m}\to R^{n}$ is an epimorphism of free modules over a nontrivial commutative ring, does it follow that $m \geq n$?  This is obviously true for vector spaces over a field, but how would one show this over just a commutative ring? -----Edit Is there any way to use the following? If $\varphi : M \to M'$ is an epimorphism of left $S$-modules and $N$ is any right $S$-module then $id_N \otimes \varphi $ is an epimorphism.
66307
Suppose $f:\mathbb{C}\rightarrow\mathbb{C}$ is an entire function. Let $g:\mathbb{C}\rightarrow\mathbb{C}$ be an entire function, which has no zeros.    I have shown that $\vert f(z) \vert \leq \vert g(z) \vert$ for all $z\in\mathbb{C}$ implies $f(z)=Cg(z)$ for some constant $C\in\mathbb{C}.$  I have to decide if this also holds if $g$ is allowed to have zeros. Help?
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I think it wants me to show that there's no bijection between the two sets.  I first tried to show that there's simply no bijection, then I realised that it doesn't work.  If I'm to show there's no bijective morphism that carries multiplication in nonzero real numbers to addition in real numbers, how am I supposed to do it? I'm thinking about doing something with 0, since it's the element of the first set but not the second set.  Thanks!
51875
What are the elements in the zeroth homotopy group? Also, why does $\pi_0(X)=0$ imply that the space is path-connected?  Thanks for the help. I find that zeroth homotopy groups are rarely discussed in literature, hence having some trouble understanding it. I do understand that the elements in $\pi_1(X)$ are loops (homotopy classes of loops), trying to see the relation to $\pi_0$.
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Prove that $[0,1]$ isn't homeomorphic to $\mathbb{R}$  My first thought is that there can not be a continuous bijection, $f$, from $[0,1]$ to $\mathbb{R}$ because a continuous function that maps $[0,1]\rightarrow \mathbb{R}$ must be bounded so there can't be a surjection. So that would then be the proof.   Though I believe I am incorrect in my thinking because a hint for the assignment says to use the intermediate value theorem. That is "Suppose $f:[a,b]\rightarrow\mathbb{R}$ is continuous. If $f(a)&lt;\delta&lt;f(b)$ or $f(b)&lt;\delta&lt;f(a)$ then $\delta=f(c)$ for some $c\in[a,b]$".   Why is my first thought wrong and how is the IVT useful?
92801
I want to ask about the homeomorphism between $\mathbb{Q}$, $\mathbb{Q}_{&gt;0}$: the rationals greater than $0$, and $\mathbb{Q}_{\geqslant 0}$: the rationals $\geqslant 0$?  For $\mathbb{Q}$ and $\mathbb{Q}_{&gt;0}$, I can use the back and forth map to have an order-isomorphism, and as a result, a homeomorphism. Is there a direct formula for the map?  It's very surprising that $\mathbb{Q}_{\geqslant 0}$ is homeomorphic to $\mathbb{Q}_{&gt;0}$ and $\mathbb{Q}$,  I don't know how to show they are homeomorphic.    Could anyone please help me with the homeomorphisms between them?
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I am learning basics of the category theory (CT).  I do understand that CT is a modern powerful framework to describe various branches of mathematics in a unified way.  I do admit that category's definition tells a lot about the whole thing and focuses on the most important properties, shadowing unnecessary details. However, the more I go forward, the more it feels like it should explicitly require a notion of morphisms equality (like it does with morphism composition). It feels so because equality or inequality of morphisms arises everywhere, since the very beginning: any commutative diagram eventually reduces to an equality of two different paths and many other things involve equality as well. Even composition is defined grounding upon an (implicit) equality!  I'll try to give an example of my concern. Consider $\mathbb{SET}$. Its objects are sets and morphisms are functions. Now, let's recall that functions are sets as well - to be more precise, a function $f : A \mapsto B$ is a subset of a $A \times B$ obeying certain properties, which I omit. Thus, one could argue that $f$ is both a morphism between objects $A, B \in Obj(\mathbb{SET})$ and an object defined as a set of pairs $\{ (a_1 \in A, b_1 \in B), ..., (a_n \in A, b_n \in B) \}$ and those are equal because both are essentially exactly the same thing.  I hope my concern is clear and I need some kind of "conceptual insight" from knowledgeable people to move forward. Please give me the right perspective.
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I am working on the following homework problem:  Let $\phi$ be an isomorphism from $\mathbb{R}^*$ to $\mathbb{R}^*$ (nonzero reals under multiplication). Show that if $r&gt;0$, then $\phi(r) &gt; 0$.  I was trying to prove this by contradiction, but I haven't been able to find any problems. I know that we must have $\phi(1) = 1$, and $\phi(-1) = -1$ no matter what, but despite playing around with those values, as well as $\pm r$ and $\pm \frac{1}{r}$, I haven't gotten any contradictions, and I'm starting to think this might be possible.  What if we defined $\phi(x)$ as: $$x\ \text{if}\ x=\pm 1 \\ -x \text{, otherwise}$$  Would this not be an isomorphism?  Any hints for this question would be much appreciated.
118524
I don't know if this is appropriate for math.stackexchange, or whether philosophy.stackexchange would have been a better bet, but I'll post it here because the content is somewhat technical.  In ZFC, we can prove that the (second-order) Peano axioms have a model, call it $\mathbb{N} = (N,0,S)$. Furthermore, $\mathbb{N}$ is unique up to isomorphism. Thus, it would seem that we've pinned down $\mathbb{N}$.  However, if ZFC is consistent, then it has some very peculiar models. In particular, it has a model $(V_0,\in_0)$ whose "native" $\mathbb{N}$, let call it $\mathbb{N}_0$, is a model for the sentence "ZFC is consistent," and another model $(V_1,\in_1)$ whose "native" $\mathbb{N}$, lets call it $\mathbb{N}_1$, is a model for the sentence "ZFC is inconsistent."  But in reality, in other words, for the "real" $\mathbb{N}$, only one of these can be sentences can be true. Furthermore, the objects $\mathbb{N}_0$ and $\mathbb{N}_1$ aren't isomorphic, since the set of sentences they satisfy are different.  So it seems to me that $\mathbb{N}$ cannot truly be pinned down. Furthermore, its categoricity is, in some sense, illusory. Is this correct, or am I missing something?
72435
No other axioms. There are $3$ sets such that:  $x\in y$; $y\in z$; $z\in x$;  These sets violate foundation.    I don't get it... For $x$, $z\in x$, so z must be the $\epsilon$-minimal member but the only member of $z$ is $y$.  $z \cap x = \{y\}\cap\{z\} = \emptyset$. None of the members of $z$ (only $y$) are members of $x$ so doesn't that satisfy definition of Foundation?  Please help this noob, thanks
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The proof I have starts of with $\;xy=0\;$ in a field.   Then $x^{-1}$ exists because it is a field.  Then $x^{-1} xy=x^{-1} 0$.  Therefore $y=0$.  But surely if an integral domain can not have any zero divisors, how can we end the proof by saying $y=0$? Surely then y is a zero divisor and hence the field is not an integral domain?
109272
I want to find all group homomorphisms $\varphi: \mathbb{Z}/ m \mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$, with $m$ and $n$ natural numbers.  Clearly $\varphi(0)=0$ since the identity in $\mathbb{Z}/m\mathbb{Z}$ must map to the identity in $\mathbb{Z}/n\mathbb{Z}$, and it follows that $m \varphi(1)=0$, so is $mx \equiv 0 \bmod n$?  This as far as I get. How can I find all the homomorphisms?
18219
I have a  basic question  Let $f$ be a $\mathcal{C}^1$  map  from  $[0,1]$ into $\mathbb{R}$ vanishing in $0$ and $1$. I want to prove that   $$\int_0^1 f'^2 \geq \pi^2 \int_0^1 f^2 $$  and to see when we have equality.
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Given some category ${\mathcal C}$, then the opposite category will consist of the same objects with the morphisms "turned around."  Given $f:A\rightarrow B$, for $A,B$ objects of ${\mathcal C}$, then in general do we have a canonical $f^{op}:B\rightarrow A$ which is induced by $f$?    Slight Motivation:  This is something that's just been bothering me recently, and I feel like I am thinking about this concept incorrectly.  To make this question slightly less vague, if we had, say, the category of rings and we had the mapping $f:{\mathbb Z}\rightarrow {\mathbb Z}/2{\mathbb Z}$ by $f$ sends even numbers to 0 and odd numbers to 1, then what is the corresponding map in the opposite category, $f^{op}:{\mathbb Z}/2{\mathbb Z}\rightarrow {\mathbb Z}$ induced by $f$?  Also, I always see this $f^{op}$ that I've mentioned just written as $f$; is this shorthand, or is there something deeper there?
80708
Please guide me in this problem. I am confused about whether its asking that having the relation $z&gt;0$ does not satisfy the order axioms.  Any help would be really appreciated.  Thanks!