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Jan
6
answered Modules over associative algebras are just special cases of “ordinary” modules over rings?
Jan
6
comment Modules over associative algebras are just special cases of “ordinary” modules over rings?
But in this case, since condition (5) is needed only to ensurethat the induced structure of $R$-module and the already present structure of $R$-module on $M$ do coincide, couldn't one simply define a module over the algebra $A$ to be simply an $A$-module: since every $A$-module has naturally the induced $R$-module structure?
Jan
4
comment Is the 1-tuple (x) = x?
@andrew Both the solutions I've described above are used, each one has its advantages: for instance the unary-product-as-the-type approach is simpler and gives a simpler notation, on the other hand if you want to use a general product operation (for instance if you need a dependent product operation) treating unary products as new types can be a better choice.
Jan
3
answered What are these diagrams called? And, what are some good *free* books/notes where I can learn about them?
Jan
3
comment Group and subgroup
Groups also have to satisfy the associativity property.
Jan
3
answered Is the 1-tuple (x) = x?
Jan
3
answered Dummit and Foote exercise verification?
Dec
31
comment Example 2.23 in Hatcher's algebraic topology: basis of homology group
Sorry for not answering ... I had no time, but I'm glad you'd completed the computations :)
Dec
28
comment Example 2.23 in Hatcher's algebraic topology: basis of homology group
@Herace The first isomorphism sends $i_n$ in $\partial i_n$, the second sends $i_{n-1}$ in either $\partial i_n$ or $-\partial i_n$ so the composite of the first isomorphism with the inverse of the second isomorphism gives an isomorphism that sends $i_n$ in $i_{n-1}$.... does this answer to your question?
Dec
27
answered Example 2.23 in Hatcher's algebraic topology: basis of homology group
Dec
22
revised Homology functors preserve coproducts
added 64 characters in body
Dec
22
revised What Does “Same Algebraic Structure” Mean?
Grammar corrections
Dec
22
answered Homology functors preserve coproducts
Dec
22
answered What Does “Same Algebraic Structure” Mean?
Dec
10
comment Deformation retract and homotopy equivalence
@grayQuant $D$ is relative to $A$ means that $D$ keep fixed the points of $A$: by definition $\forall a \in A, t \in I$ we have $D(a,t)=a$. Since $D(x,1)=i\circ r(x)$ we have that for $x=a \in A$ it must be $i \circ r(a) = D(a,1)=a$ and since $i$ is an embedding (that is $i(x)=x$ for every $x \in A$) we have that $r(a)=a$ (this proves that $r$ fixes the points of $A$). Hope this helps.
Oct
31
comment In the category of rings, what is an example of an epimorphism that is not a retraction?
@HaloLikeAHat there are many indeed: consider for instance the canonical projection $\pi \colon \mathbb Z \to \mathbb Z/2\mathbb Z$, which is clearly surjective. This mapping has no left inverse, otherwise $\mathbb Z/2\mathbb Z$ should be (isomorphic) to a subring(and so to a subgroup) of $\mathbb Z$ which is not possible.
Oct
31
answered Cofibration and retraction
Oct
31
answered In the category of rings, what is an example of an epimorphism that is not a retraction?
Oct
19
answered Creating DFA to prove closure properties
Sep
1
answered Group Action as permutations