### Induced homomorphisms

So far we have associated the fundamental group to a based space, i.e., a space with a given based point, $(X, x_0)$ and computed this in a few cases. We shall now associate to maps between based spaces $f: (X, x_0)\to (Y, y_0)$ homomorphisms. Here the map $f: (X, x_0)\to (Y, y_0)$ is a map from $X$ to $Y$ with $f(x_0) = y_0$.

We define $f_*: \pi_1(X, x_0) \to \pi_1(Y, y_0)$ by

for $\alpha: [0, 1]\to X$ with $\alpha(0)=\alpha(1)=x_0$.

It is easy to see that

- $f_*$ is well-defined.
- $f_*$ is a homomorphism.
- for an identity map $id$, $id_*$ is the identity homomorphism.
- $(g\circ f)_* = g_*\circ f_*$.

These let us translate topological problems to algebraic ones to get some nice results.

### No retraction theorem

**Theorem**: There is no map $\rho: D^2 \to S^1$ so that for all $x\in S^1$, $\rho(x)=x$.

Suppose such a map $\rho: (D^2, 1)\to (S^1, 1)$ exists, if $i: (S^1, 1)\to (D^2, 1)$ denotes the inclusion, then as $\rho\circ i$ is the identity map, it follows that $\rho_* \circ i_* : \pi_1(S^1, 1) \to \pi_1(S^1, 1)$ is the identity. But $\pi_1(S^1, 1) = \mathbb{Z}$, while $\pi_1(D^2, 1)$ is trivial. The latter shows that the homomorphism $\rho_* $ is $0$, while the former shows that the composition $\rho_* \circ i_* $ is a non-zero homomorphism. This gives a contradiction.