# Ayman Hourieh

Dublin, Ireland

 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-ModulesThe 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 productMore 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 pointFixed 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 coveredCareful, 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 coveredWhile 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.