Apr
18
revised Let $A$ be the annulus in $\mathbb{C}$, what is the space $A/{\sim}$ generated by identifying all points on the "inner circle" with each other?
Fix spacing around \sim
Apr
18
comment Motivation For Tensor Product of R-Modules
The next section in Dummit & Foote introduces projective, injective and flat modules. It also proves the tensor-hom adjunction. These concepts are based on the tensor product, and are crucial in homological algebra, commutative algebra, algebraic topology and algebraic geometry. For example, see the Künneth theorem; it allows you to compute the homology groups of a product space via the tensor product. There are many, many more examples. Just keep going. :)
Apr
18
revised Module structure of base extension via tensor product
More searchable title
Apr
17
answered Relative homology $H_n(S^2,S^0)$, or other examples
Apr
17
comment Let $X$ be a normal space then there exists a continuous map $f : X → [0, 1]$ such that $f^{−1} (0) = A$ and $f^{−1} (1) = B$
possible duplicate of [Let $X$ be a normal space then there exists a continuous map $f : X → [0, 1]$ such that $f^{−1} (0) = A$](math.stackexchange.com/questions/1239003/…)
Apr
17
revised Quotient topology by identifying the boundary of a circle as one point
Fixed typo
Apr
17
answered Quotient topology by identifying the boundary of a circle as one point
Apr
17
comment Let $R$ be a ring with 1 and N be a submodule of R-module M. If $M$ is free of finite rank, is $M/N$ necessarily free of finite rank?
A non-empty direct sum of copies of $\mathbb Z$ must be infinite, right? Since $\mathbb Z / 2 \mathbb Z$ is finite, it cannot be free.
Apr
17
comment Let $R$ be a ring with 1 and N be a submodule of R-module M. If $M$ is free of finite rank, is $M/N$ necessarily free of finite rank?
Looks good. You can also start with $\mathbb Z$ and note that $\mathbb Z / 2 \mathbb Z$ cannot be a free $\mathbb Z$-module since it's finite.
Apr
16
comment Field $K (x)$ of rational functions with coefficients from $K$, if $f\in K(x)$, then $f^2 \neq x^2-1$
Your counterexample works. If this is indeed how the question is presented, then it's incorrect.
Apr
16
comment Field $K (x)$ of rational functions with coefficients from $K$, if $f\in K(x)$, then $f^2 \neq x^2-1$
I edited your question to correct this. Does it look good now?
Apr
16
revised Field $K (x)$ of rational functions with coefficients from $K$, if $f\in K(x)$, then $f^2 \neq x^2-1$
deleted 1 character in body; edited title
Apr
16
awarded Good Answer
Apr
16
comment Prove that a non-empty subset of an open set which is evenly covered is evenly covered
Careful, it's $V_\alpha \cap p^{-1}(W)$ that is homeomorphic to $W$, not $\bigcup V_\alpha \cap p^{-1}(W)$. Stefan's answer below elaborates on this idea.
Apr
16
comment Prove that a non-empty subset of an open set which is evenly covered is evenly covered
While not universal, the term is also used in Spanier's Algebraic Topology and Lee's Introduction to Topological Manifolds (among others). @Nescrio Did you try just restricting $U$'s cover to $W \subset U$? If $V_\alpha$ is homeomorphic to $U$ via $p$, then $V_\alpha \cap p^{-1}(W)$ is homeomorphic to $W$ via $p$.
Apr
15
comment The inclusion $\mathbb Z \to \mathbb Q$ is an epimorphism
@mdlt Done. ${}$
Apr
15
answered The inclusion $\mathbb Z \to \mathbb Q$ is an epimorphism
Apr
15
comment Let $\phi:R[X] \rightarrow S[X]$ be a unital ring homomorphism. Prove if $f(x) \in R[X]$ is reducible, then $\phi(f(x))$ is reducible.
How do you define 'reducible'? If you require it to have a non-trivial factorization, then this isn't true in general. Let $R = S = k$ be a field, and consider $\varphi : k[x] \to k[x]$, $f(x) \mapsto f(0)$. Then $x^2 - 1$ is reducible, but its image is not.
Apr
14
revised Degree of field extension $F(x) / F(x^2 + 1 / x^2)$
edited tags; edited title
Apr
14
comment Degree of field extension $F(x) / F(x^2 + 1 / x^2)$
No problem at all, dear @Georges. I always appreciate your comments and feedback. Thank you for linking to my answer. I think I'll mark that question as a duplicate of this.
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