In a lecture in Symplectic Geometry, the professor said that $\pi_1(U(n))=\mathbb Z$ because $\det:U(n)\to S^1$ is a continuous map (in fact, a Lie group morphism) which induces an isomorphism $\det_\ast:\pi_1(U(n))\to\pi_1(S^1)=\mathbb Z$.
To prove this, I first recalled $\det_\ast[\gamma]=[\det\circ\gamma]$, then laid out the assumptions and the thesis.
Assumption We have a path in $U(n)$ which, via $\det$, gives a contractible path in $S^1$. In other words, we have a continuous $\gamma:[0,1]\to U(n)$ and a continuous $f:S^1\times[0,1]\to S^1$ such that, for any $t$:
$$f(\det\gamma(t),0)=\det\gamma(t),\qquad f(\det\gamma(t),1)=\det\gamma(0).$$
Thesis The path $\gamma$ is contractible, that is there exists a continuous $\tilde f:U(n)\times[0,1]\to U(n)$ such that, for any $t$:
$$\tilde f(\gamma(t),0)=\gamma(t),\qquad\tilde f(\gamma(t),1)=\gamma(0).$$
And now I am wonderfully stuck. How do I deduce the thesis from the assumption? And first of all, this is correctly proving that $\det_\ast$ is an isomorphism, right?
There is certainly a nicer proof, but in the meantime this one should do.
Let $F_k$ be the space of all $k$-tuples of unit and mutually orthogonal vectors in $\mathbb C^n$, so that $F_n=U(n)$ and $F_0=$point. Notice that for every $k$ the map forgetting the last vector $F_{k+1}\to F_{k}$ is a fibre bundle with the fibre $S^{2(n-k)-1}$ (unit vectors in $\mathbb C^{n-k}$), and so by the long homotopy sequence we have $\pi_1(F_k)=\pi_1(F_{k-1})$ for every $k<n$ (when $\pi_1(S^{2(n-k)-1})=0$). Things change for $k=n-1$ - the homotopy sequence now says that $\mathbb Z=\pi_1(S^1)\to\pi_1(F_n=U(n))$ is surjective. To see that the map is an iso and that det gives its inverse just notice that the composition $S^1\to U(n)\to S^1$, where the second map is the determinant, is the identity (the map $S^1\to U(n)$ maps $a$ to the matrix containing $a$ in the bottom-right corner, otherwise with $1$'s on the diagonal and $0$'s off diagonal).