Suppose a space $X$ is obtained from a path connected space $Y$ by attaching $n$ cell ($n \geq 3$).Show that $\pi_1(X) \cong \pi_1(Y)$.
I am trying to use the Van Kampen theorem to prove this but i am unable to write it clearly. Could someone give some idea to prove this?
Usually 'attaching' means the cell $C$ intersects $X$ over some well-behaved set, for example the singleton $\{c\}$. We proceed under that assumption.
We have the two inclusion maps $f : \{c\} \to X$ and $g \colon \{c\} \to C$.
We also have three fundamental groups $\pi(X), \pi(C)$ and $\pi(\{c\})$.
Exercise: The groups $\pi(C)$ and $\pi(\{c\})$ are particularly simple. What do they look like?
The two maps induce homomorphisms between fundamental groups (with the same names).
$f : \pi(\{c\}) \to \pi(X)$ and $g : \pi(\{c\}) \to \pi(C)$
Exercise: What do these homomorphisms look like?
The van Kampen theorem says that once we know what the three groups and two homomorphisms are, we are free to forget about the topology entirely!
All we need to worry about is the mapping diagram $\pi(X) \leftarrow \pi(\{c\}) \to \pi(C)$.
The theorem says $\pi(X \cup C) = \pi(Y)$ is the pushout of that diagram. The definition of the pushout is rather long, and computing particular pushouts is often even longer.
Exercise: But in this case the groups and homomorphisms (see previous exercises) are straightforward enough that the pushout can only be $\pi(X)$.