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Could someone supply me a precise reference to the computation of all homotopy groups of infinite real projective space and infinite complex projective space?

Najib Idrissi
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    I'm sure this is done in chapter 4 of Hatcher. In any case, you have fiber bundles $S^0 \to S^\infty \to \Bbb{RP}^\infty$ and $S^1 \to S^\infty \to \Bbb{CP}^\infty$, and this follows from the long exact sequence of homotopy groups of a fibration. (In the first case it's better to think of $\Bbb{RP}^\infty$ as a quotient of $S^\infty$ by a free action of $\Bbb Z/2$.) –  Sep 28 '15 at 16:15

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Because $S^\infty$ is contractible, and we have a fiber bundles $S^1 \to S^\infty \to \Bbb{CP}^\infty$, the long exact sequence of homotopy groups of a pair shows that its only nonvanishing homotopy group is $\pi_2(\Bbb{CP}^\infty) = \Bbb Z$. See Hatcher, 4.50. Example 4.44 is a construction of this bundle.

$\Bbb{RP}^\infty$ is more elementary: if $\tilde M$ is a cover of $M$, then $\pi_i(\tilde M) = \pi_i(M)$ for all $i>1$. So because the contractible $S^\infty$ double covers $\Bbb{RP}^\infty$, it has fundamental group $\Bbb Z/2$ and no other nonzero homotopy groups. See Hatcher, 1B.3.

Balarka Sen
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    Another way to phrase these is (in terms of loop spaces): $\Omega\mathbb{CP}^\infty\simeq S^1$ and $\Omega\mathbb{RP}^\infty\simeq\mathbb Z/2$, or (in terms of classifying spaces) $\mathbb{CP}^\infty\simeq BS^1$, $\mathbb{RP}^\infty\simeq B\mathbb Z/2$. In general, it is known that $\Omega BG\simeq G$. – Kenta S Apr 16 '22 at 02:40