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No, this is false. According to the Sullivan Conjecture (Miller's Theorem), $\mathrm{map}_*(B\mathbb{Z}/p, S^n) \sim *$ for all $n$, which means $$ [\Sigma^n B\mathbb{Z}/p, S^k] = * $$ for all $n$. …

Let $f:S^n\to C$ and let $K$ be its image. If $n>1$ then the Theorem implies $f\simeq *$. EDIT: Clearly this is over-simple and wrong, as Omar Antolín-Camarena has pointed out. In fact, it looks li …

To define the relative homotopy groups of a pair $(X, A)$, let $i:A\to X$ be the inclusion, and write $F_i$ for its homotopy fiber. Then $$ \pi_n(X, A) = \pi_{n-1}(F_i). $$ In your examples, the in …

The dumb answer is infinite projective space, which has zero Betti numbers as classically computed (over $\mathbb{Q}$). But what you really want is very small cohomology and very large L-S category. …

You can think of a simplex as a finite ordered list (i.e., the vertices). The simplices of its barycentric subdivision are the lists of subsets of the first list, ordered by inclusion.

They are clearly not homotopy equivalent, and they have different homotopy groups. The nontorus is the cofiber of any injective map from a discrete $3$-point set to $S^2$, which is homotopy equival …

The existence of the maps $g_i$ is only possible if $G_i$ is abelian, and so what you are missing is that each $G_i$ must be abelian. It's been a while since I looked at that book, but it very well …

The cup product is a gussied-up version of the map induced by the reduced diagonal map $\bar \Delta$, which is the composite of the ordinary diagonal $\Delta: X\to X\times X$ and the quotient map $q: …

You can certainly embed the real line, with perpendicular circles at the integer points, into $\mathbb{R}^3$, and this is homotopy equivalent to the wedge you want. Let $W$ be the expanding `inverse …

If you accept the Whitehead theorem, then simple obstruction theory (without cohomology) determines the homotopy type of spaces with homotopy groups in only one dimension. That is, you can easily sh …

Every map from a sphere to a finite connected CW complex is homotopic to a surjective one. Collapse one half of the sphere to a line segment, use your original map on the sphere created by the collap …

You can build a weak homotopy equivalence $K \to X$ from a CW complex with $2$-skeleton $K_2 = *$. Then cellular homology gives the result, provided your homology respects weak equivalence.

If $X$ is co-H, then $\pi_1(X)$ must be a free group. If $X$ is H, then $\pi_1(X)$ must be abelian. The only free group that is abelian is $\mathbb{Z}$. Argument for the first assertion: The co-H s …

Sure. The basic point is that for simply-connected spaces, you can determine the connectivity of a map by looking at the connectivity of the cofiber instead of the connectivity of the fiber. In hom …

Write $E_A = p^{-1}(A)$. Under mild conditions, the inclusion $$ (E_A \times I)\cup (E \times \{ 0\}) \to E\times I $$ will be a cofibration and a (weak) homotopy equivalence. Now set up the liftin …