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 1d comment If an isomorphism can be expressed as a composition of morphisms, what can we say about it's components? Think of the counterexample in $\mathrm{Ab}$ as a counterexample in $\mathrm{Set}$. 1d comment If an isomorphism can be expressed as a composition of morphisms, what can we say about it's components? We cannot. Think of categories with a null object, like the category of abelian groups. (And it doesn’t work for most concrete categories like $\mathrm{Set}$ itself.) 1d comment $\dim (W_{1} \cap W_{2}) = \dim W_{1}$ implies $W_{1} \subset W_{2}$? I didn’t mean to question your competence on linear algebra. To me it seemed like the problem was meant to exercise how to handle bases and base lengths and reason with them. From this point of view, I merely found “no need to reason about bases” to be false. That the questioner accepted the hint-answer, on the other hand, speaks for your interpretation. 2d comment $\dim (W_{1} \cap W_{2}) = \dim W_{1}$ implies $W_{1} \subset W_{2}$? I think there is very much need to reason about bases. The statement doesn’t hold if you replace vector spaces with free modules and dimensions with rank, or if you go to the infinite-dimensional. You strongly use the fact that bases of finitely generated vector spaces have a unique length and are characterized by being maximal linearly independent systems. The answer you have given just subsumes all that in the step “If in addition it has the same (finite) dimension as $W_1$, then it must be all of $W_1$”. 2d comment $\dim (W_{1} \cap W_{2}) = \dim W_{1}$ implies $W_{1} \subset W_{2}$? You mean $β ⊂ W_1$. Apr23 comment Prove compact metric spaces $X$ and $Y$ are isomorphic given these conditions What is $X(Y)$? Apr21 comment Direct sum of two non-zero $R$-modules (By the way, both mistakes can easily be avoided by following the habit of not considering the direct sum “$\oplus$” to be same as the direct product of rings “$×$”, as there is no direct sum of rings. Carrying this distinction over to notation really alerts one to these kind of subtleties.) Apr21 comment Direct sum of two non-zero $R$-modules There might be a litte flaw with the first part of your answer. You assume $R \cong M \oplus M'$ for some $R$-modules $M$, $M'$, but then you use a multiplication $(m,0)·(0,m') = 0$ in $M \oplus M'$ to conclude that $R$ has zero divisors. There are two problems with that: 1. As $R$-modules $M$ and $M'$ do not carry any natural multiplication. 2. Even if they do (say by arising as quotients $M = R/I$, $M' = R/J$ for some ideals $I$, $J$ of $R$), the isomorphism $R \cong M \oplus M'$ isn’t guaranteed to be a ring isomorphism as well (if $M \oplus M'$ is seen as the ring $M × M'$). Apr19 comment How to prove that in this context epimorphisms are 'surjective' Not sure if that is illegal, but I upvoted solely for “poor me”. (Update: And now that I read the entire question, I don’t feel guilty about it.) Apr17 comment Proving every element of $\mathbb{F}[x]/(p(x))$ can be expressed uniquely @DH. Best practice: Answer it yourself, elaborating a bit, accept your own answer. Else, I’d just leave it. Apr17 comment Proving every element of $\mathbb{F}[x]/(p(x))$ can be expressed uniquely @DH. Yes, seems legit. Apr17 comment Proving every element of $\mathbb{F}[x]/(p(x))$ can be expressed uniquely @DH. Well, maybe you can say at least something about the degree of $a - a_1$, even if you can’t know the degree of $a + b$ for polynomials $a$ and $b$ for sure if you only know their degrees. (You can’t know it because, after all, if $a = X^n + X^{n-1} + … + 1$ and $b = - (X^n + X^{n-1} + … + X^{k+1})$ for some $k ∈ \{0,…,n\}$, then $a + b = X^k + X^{k-1} + … + 1$ has degree $k$ which could be anything. Well, of course anything between $0$ and $n$ …) Apr17 comment Proving every element of $\mathbb{F}[x]/(p(x))$ can be expressed uniquely What’s the degree of $a(x) - a_1(x)$, what’s the degree of $q(x)·p(x)$? Put differently: Can you bound the former from above, the latter from below? Apr17 comment Let $X$ be a normal space then there exists a continuous map $f : X → [0, 1]$ such that $f^{−1} (0) = A$ You probably need to say a thing or two about why $f$ still is continuous. Apr17 comment Values of the Herbrand quotient Why would anyone downvote this question? Apr17 comment Let $X$ be a normal space then there exists a continuous map $f : X → [0, 1]$ such that $f^{−1} (0) = A$ @5xum Is it? The property doesn’t say that the closed sets in $X$ are exactly countable intersections of open sets, so there might be countable intersections of open sets (such as trivial ones) that aren’t closed, right? Apr11 comment I have a question about the multiplicative inverse in any field. It should. If you tell us the field axioms you are using, I’m sure we can point out which one assures $0 ≠ 1$. Anyway, the only ring $R$ with one in which $0 = 1$ is $R = 0$, since $∀x ∈ R\colon x = x·1 = x·0 = 0$. Apr11 comment I have a question about the multiplicative inverse in any field. If $α$ was a multiplicative inverse of $0$, then $α·0 = 1$, but also $α·0 = 0$ (which is true for all elements $a$ of the field since $a·0 = a·(0+0) = a·0 + a·0$ from which you can additively cancel $a·0$ (by adding $-a·0$ to both sides)). Hence $0 = 1$ which is not allowed by the very definition of a field. Apr11 comment about hyperplanes in finite fields Hyperplanes are vector spaces and if $U ⊂ \mathbf F_q^k$ is a ($k-2$-dimensional) proper subspace and $x \notin U$, then $U ∪ \{x\}$ just won’t be a vector space (think of all the sums $x + u$ with $u ∈ U$ you are missing). Apr11 comment Definition of split of exact sequence @ringwith1 You are thinking of this: If $M$ is an abelian group, then for subgroups $A ⊂ M$ and $B ⊂ M$ it is equivalent that $M$ splits as an internal direct sum $M = A \oplus B$ if and only if for every $m ∈ M$ there are unique $a ∈ A$, $b ∈ B$ such that $m = a + b$. In your case, the latter part of the equivalence may be satisfied in some sense, but $ℤ/nℤ$ $(=B)$ just isn’t a subgroup of $ℤ$ $(=M)$ (and even not so if you regard $ℤ/nℤ$ as the subset $\{0,…,n-1\}$ of $ℤ$, because then the addition on $ℤ/nℤ$ is not inherited from the addition in $ℤ$).