homology group of adjunction space

Question

I start to study homology theory and i want to understand homology groups of adjunction space

In this picture i can't see $V$ deformation retracts to $X$ neither intuitively nor explicitly help please intuitively and with explicit formula of deformation retraction

Answer 1

Note that $E^k \setminus \{p\}$ deformation retracts to $S^{k-1}$ by $r : E^k \setminus \{p\} \to S^{k-1}, \; x \mapsto \frac x {\| x \|}$. Therefore, $\mathrm{id} \sqcup _f r : X \sqcup _f (E^k \setminus \{ p \}) = V \to X \sqcup _f S^{k-1} \simeq X$ is a retract of $V$ to $X$.

To clarify why $X \sqcup _f S^{k-1} \simeq X$ let us show that, in general, if $f : S \to X$ is continuous then $X \sqcup _f S \simeq X$. Let $i : X \to X \sqcup S$ and $\pi : X \sqcup S \to X \sqcup _f S$ be the canonical injection and, respectively, projection (both continuous by definition, so $\pi i : X \to X \sqcup _f S$ will also be continuous).

Note that $\pi i$ is injective: $\pi i (x) = \pi i (x')$ means $\pi (x) = \pi (x')$, so $x$ is equivalent to $x'$ modulo $f$. But the only pairs of points from $X \sqcup S$ that are equivalent must be of the form $(x,x)$, $(s,s)$ and $(f(s),s)$. Since $x,x' \in X$, then $x = x'$.

Note that $\pi i$ is surjective: pick $z \in X \sqcup _f S$; since $\pi$ is surjective, there exist $p \in X \sqcup S$ with $\pi (p) = z$. If $p \in X$, then $\pi i (p) = \pi (p) = z$. If $p \in S$, then $f(p) \in X$ and, more, $p$ and $f(p)$ will be equivalent modulo $f$, so $\pi i (f(p)) = z$.

Thus, $\pi i$ is bijective and continuous. To construct a continuous inverse $u$, note that the adjunction space is a particular case of a pushout, so we shall use the universality property of the pushout. In the diagram on Wikipedia, choose $Z = Y = S$, $g = \mathrm{id}_S$, $P = X \sqcup _f S$, $i_1 = \pi i$, $Q = X$, $j_1 = \mathrm{id}_X$ (you won't need $i_2$ and $j_2$). Then, the universal property guarantees the existence of a continuous $u$ (unique up to homeomorphism) such that $u \circ \pi i = \mathrm{id}_X$. Since $\pi i$ has been proven invertible, $u$ will also be a right inverse, so $u = (\pi i)^{-1}$.

Therefore, we have just proved that $\pi i$ is a homeomorphism.

Answer 2

There is a more general notion of a mapping cone. That is, an adjunction space corresponding to $f:S^{k-1} \to X$ is the same as a mapping cone $C_f$.

For those you find plenty of answers which use the same strategy, hence you can build your intuition and see formulas by considering e.g. https://en.wikipedia.org/wiki/Mapping_cone_(topology), Homology of mapping cone, standard topology books and google.

Note that this result for the homology groups leads also to an analagous notion for general chain complexes https://en.wikipedia.org/wiki/Mapping_cone_(homological_algebra).