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 8h comment Correct definition of model category @QiaochuYuan I just looked it up. Hovey apparently distinguishes between a category with a Model structure and a Model category. The latter is a category having all small limits/colimits, equipped with a Model structure. 8h revised In a model category, is the full subcategory of fibrant objects a reflective subcategory? added 23 characters in body 8h comment In a model category, is the full subcategory of fibrant objects a reflective subcategory? @DanielGerigk On the other hand, according to Hovey, model categories do come equipped with strong factorization systems. Hence, I believe the validity of the above statement is a matter of semantics. 9h awarded Student 9h comment Correct definition of model category @Kevin Carlson Yes, I think he says that in all the examples he considers, the factorization can be made functorial. I also know he uses the fact frequently. I don't think a weak factorization system on a model category always implies the existence of a functorial factorization. I'm looking for examples where statements cannot be proved without using functoriality. 10h asked Correct definition of model category 2d answered In a model category, is the full subcategory of fibrant objects a reflective subcategory? Jan 15 revised Spectrum $\sigma(T)$ of $T:l^1 \to l^1$ given by $T((a_j))=\left( \sum_{j=2}^{\infty} a_j \right) e_1 + \sum_{j=2}^{\infty} a_{j-1} e_j$ added 737 characters in body Jan 15 comment Spectrum $\sigma(T)$ of $T:l^1 \to l^1$ given by $T((a_j))=\left( \sum_{j=2}^{\infty} a_j \right) e_1 + \sum_{j=2}^{\infty} a_{j-1} e_j$ ok I've edited the post. Jan 15 revised Spectrum $\sigma(T)$ of $T:l^1 \to l^1$ given by $T((a_j))=\left( \sum_{j=2}^{\infty} a_j \right) e_1 + \sum_{j=2}^{\infty} a_{j-1} e_j$ added 737 characters in body Jan 15 comment Spectrum $\sigma(T)$ of $T:l^1 \to l^1$ given by $T((a_j))=\left( \sum_{j=2}^{\infty} a_j \right) e_1 + \sum_{j=2}^{\infty} a_{j-1} e_j$ @Kika Maybe I'm wrong, but I thought you can just calculate the eigenvalues of the adjoint. I think the conjugates give you the rest of the spectrum. Jan 15 revised Spectrum $\sigma(T)$ of $T:l^1 \to l^1$ given by $T((a_j))=\left( \sum_{j=2}^{\infty} a_j \right) e_1 + \sum_{j=2}^{\infty} a_{j-1} e_j$ added 40 characters in body Jan 15 answered Spectrum $\sigma(T)$ of $T:l^1 \to l^1$ given by $T((a_j))=\left( \sum_{j=2}^{\infty} a_j \right) e_1 + \sum_{j=2}^{\infty} a_{j-1} e_j$ Jan 15 comment Spectrum $\sigma(T)$ of $T:l^1 \to l^1$ given by $T((a_j))=\left( \sum_{j=2}^{\infty} a_j \right) e_1 + \sum_{j=2}^{\infty} a_{j-1} e_j$ What exactly is the operator? If your summing over $j$'s, $T((a_j))$ is not a sequence. Dec 15 awarded Necromancer Dec 8 revised Can curvature be defined in Topos Theory? edited body Dec 6 awarded Revival Dec 6 revised Can curvature be defined in Topos Theory? added 12 characters in body Dec 6 answered Can curvature be defined in Topos Theory? Dec 5 awarded Yearling