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If the spectrum $X$ is $r$-connected, then the map $\Sigma^\infty\Omega^\infty X \to X$ is $(2r+2)$-connected.

Here's a sketch: apply the functor $\Omega^\infty$ to get the map of spaces $$ Q(\Omega^\infty X) \to \Omega^\infty X $$ where $Q = \Omega^\infty\Sigma^\infty$. It will be enough to identify the connectivity of the latter.

This map has a section so it is surjective on homotopy in all degrees. In particular your map of spectra is surjective on homotopy in all degrees.

It's enough to compute the degree of injectivity. Let $S^j \to Q(\Omega^\infty X)$ be a map, $j \le 2r+1$. Suppose that its image in $\pi_j(\Omega^\infty X)$ is trivial.

By Freudenthal, the map $$ \pi_j(\Omega^\infty X) \to \pi_j(Q(\Omega^\infty X)) $$ is surjective in this degree ($j \le 2r+1$). So our map desuspends to give a map $S^j \to \Omega^\infty X$. But the composite $$ \pi_j(\Omega^\infty X) \to \pi_j(Q(\Omega^\infty X))\to \pi_j(\Omega^\infty X) $$ is the identity. This means that our desuspension is trivial, hence its iterated suspensions are as well.

Here is the essence of the above argument: suppose $A\to B\to C$ are maps of spaces in which $A\to B$ is $k$-connected and $A\to C$ is $(k+1)$-connected. Then $B\to C$ is $(k+1)$-connected as well. Apply this to $$ \Omega^\infty X \to Q(\Omega^\infty X) \to \Omega^\infty X . $$

If the spectrum $X$ is $r$-connected, then the map $\Sigma^\infty\Omega^\infty X \to X$ is $(2r+2)$-connected.

Here's a sketch: apply the functor $\Omega^\infty$ to get the map of spaces $$ Q(\Omega^\infty X) \to \Omega^\infty X $$ where $Q = \Omega^\infty\Sigma^\infty$. It will be enough to identify the connectivity of the latter.

This map has a section so it is surjective on homotopy in all degrees. In particular your map of spectra is surjective on homotopy in all degrees.

It's enough to compute the degree of injectivity. Let $S^j \to Q(\Omega^\infty X)$ be a map, $j \le 2r+1$. Suppose that its image in $\pi_j(\Omega^\infty X)$ is trivial.

By Freudenthal, the map $$ \pi_j(\Omega^\infty X) \to \pi_j(Q(\Omega^\infty X)) $$ is surjective in this degree ($j \le 2r+1$). So our map desuspends to give a map $S^j \to \Omega^\infty X$. But the composite $$ \pi_j(\Omega^\infty X) \to \pi_j(Q(\Omega^\infty X))\to \pi_j(\Omega^\infty X) $$ is the identity. This means that our desuspension is trivial, hence its iterated suspensions are as well.

Here is the essence of the above argument: suppose $A\to B\to C$ are maps of based spaces in which $A\to B$ is $k$-connected and $A\to C$ is $(k+1)$-connected. Then $B\to C$ is $(k+1)$-connected as well. Apply this to $$ \Omega^\infty X \to Q(\Omega^\infty X) \to \Omega^\infty X . $$

If the spectrum $X$ is $r$-connected, then the map $\Sigma^\infty\Omega^\infty X \to X$ is $(2r+2)$-connected.

Here's a sketch: apply the functor $\Omega^\infty$ to get the map of spaces $$ Q(\Omega^\infty X) \to \Omega^\infty X $$ where $Q = \Omega^\infty\Sigma^\infty$. It will be enough to identify the connectivity of the latter.

This map has a section so it is surjective on homotopy in all degrees. In particular your map of spectra is surjective on homotopy in all degrees.

It's enough to compute the degree of injectivity. Let $S^j \to Q(\Omega^\infty X)$ be a map, $j \le 2r+1$. Suppose that its image in $\pi_j(\Omega^\infty X)$ is trivial.

By Freudenthal, the map $$ \pi_j(\Omega^\infty X) \to \pi_j(Q(\Omega^\infty X)) $$ is surjective in this degree ($j \le 2r+1$). So our map desuspends to give a map $S^j \to \Omega^\infty X$. But the composite $$ \pi_j(\Omega^\infty X) \to \pi_j(Q(\Omega^\infty X))\to \pi_j(\Omega^\infty X) $$ is the identity. This means that our desuspension is trivial, hence its iterated suspensions are as well.

If the spectrum $X$ is $r$-connected, then the map $\Sigma^\infty\Omega^\infty X \to X$ is $(2r+2)$-connected.

Here's a sketch: apply the functor $\Omega^\infty$ to get the map of spaces $$ Q(\Omega^\infty X) \to \Omega^\infty X $$ where $Q = \Omega^\infty\Sigma^\infty$. It will be enough to identify the connectivity of the latter.

This map has a section so it is surjective on homotopy in all degrees. In particular your map of spectra is surjective on homotopy in all degrees.

It's enough to compute the degree of injectivity. Let $S^j \to Q(\Omega^\infty X)$ be a map, $j \le 2r+1$. Suppose that its image in $\pi_j(\Omega^\infty X)$ is trivial.

By Freudenthal, the map $$ \pi_j(\Omega^\infty X) \to \pi_j(Q(\Omega^\infty X)) $$ is surjective in this degree ($j \le 2r+1$). So our map desuspends to give a map $S^j \to \Omega^\infty X$. But the composite $$ \pi_j(\Omega^\infty X) \to \pi_j(Q(\Omega^\infty X))\to \pi_j(\Omega^\infty X) $$ is the identity. This means that our desuspension is trivial, hence its iterated suspensions are as well.

Here is the essence of the above argument: suppose $A\to B\to C$ are maps of spaces in which $A\to B$ is $k$-connected and $A\to C$ is $(k+1)$-connected. Then $B\to C$ is $(k+1)$-connected as well. Apply this to $$ \Omega^\infty X \to Q(\Omega^\infty X) \to \Omega^\infty X . $$

If the spectrum $X$ is $r$-connected, then the map $\Sigma^\infty\Omega^\infty X \to X$ is $(2r+2)$-connected.

Here's a sketch: apply the functor $\Omega^\infty$ to get the map of spaces $$ Q(\Omega^\infty X) \to \Omega^\infty X $$ It will be enough to identify the connectivity of the latter.

This map has a section so it is surjective on homotopy in all degrees. In particular your map of spectra is surjective on homotopy in all degrees.

It's enough to compute the degree of injectivity. Let $S^j \to Q(\Omega^\infty X)$ be a map, $j \le 2r+1$. Suppose that its image in $\pi_j(\Omega^\infty X)$ is trivial.

By Freudenthal, the map $$ \pi_j(\Omega^\infty X) \to \pi_j(Q(\Omega^\infty X)) $$ is surjective in this degree ($j \le 2r+1$). So our map desuspends to give a map $S^j \to \Omega^\infty X$. But the composite $$ \pi_j(\Omega^\infty X) \to \pi_j(Q(\Omega^\infty X))\to \pi_j(\Omega^\infty X) $$ is the identity. This means that our desuspension is trivial, hence its iterated suspensions are as well.

If the spectrum $X$ is $r$-connected, then the map $\Sigma^\infty\Omega^\infty X \to X$ is $(2r+2)$-connected.

Here's a sketch: apply the functor $\Omega^\infty$ to get the map of spaces $$ Q(\Omega^\infty X) \to \Omega^\infty X $$ where $Q = \Omega^\infty\Sigma^\infty$. It will be enough to identify the connectivity of the latter.

This map has a section so it is surjective on homotopy in all degrees. In particular your map of spectra is surjective on homotopy in all degrees.

It's enough to compute the degree of injectivity. Let $S^j \to Q(\Omega^\infty X)$ be a map, $j \le 2r+1$. Suppose that its image in $\pi_j(\Omega^\infty X)$ is trivial.

By Freudenthal, the map $$ \pi_j(\Omega^\infty X) \to \pi_j(Q(\Omega^\infty X)) $$ is surjective in this degree ($j \le 2r+1$). So our map desuspends to give a map $S^j \to \Omega^\infty X$. But the composite $$ \pi_j(\Omega^\infty X) \to \pi_j(Q(\Omega^\infty X))\to \pi_j(\Omega^\infty X) $$ is the identity. This means that our desuspension is trivial, hence its iterated suspensions are as well.