On $\pi_n$ commuting with certain functors if $f:X\to Y$ is continuous and injective, then $\pi_n(f(X))=f_*(\pi_n(X))$
This stackexchange poster asked an interesting but naïve question. The category of pointed topological spaces has $0$-objects and thus $0$-morphisms. What if we instead of considering f(X) = im(f) = ker(coker(f)) we define hocoker as the homotopy colimit of the diagram (how do you draw two arrows on top of eachother? Just imagine that there are two arrows on top of eachother) $$X\xrightarrow{0,f} Y$$ and hoker as the homotopy limit of the same diagram. Then if we define hoim(f) = hoker(hocoker(f)) we can rephrase the original question.
Is it true that $f_*(\pi_n(X))=\pi_n($hoim$(f))$ but now we drop the injectivity condition on $f$ since every $f$ is pointwise equivalent to an injective cofibration $X \rightarrow M_f$
The original poster was incorrect essentially because a pointwise homotopy equivalence between diagrams doesn't necessarily induce a homotopy equivalence of colimits or limits. But this is the case for the holim and hocolim functors. For a concrete example, consider the map $I \xrightarrow {1/2} I$ which is a homotopy equivalence and the quotient map $I \xrightarrow f S^1$. Then defining $g = f \circ 1/2$. Then g and f are obviosuly pointwise homotopy equivalent but im$(f)$ and im$(g)$ are not homotopy equivalent.
This is more of an open ended question, please come with your own suggestions of changes or whatever. Thanks in advance.
 A: The homotopy equaliser of
$$X\stackrel{f,\ast}{\rightrightarrows}Y$$
is just the cofibre $C_f$ of $f$. Similarly, the homotopy equaliser of
$$Y\stackrel{k,\ast}{\rightrightarrows}C_f$$
is just the homotopy fibre $F_k$ of $k=k_f$. Clearly it is very unlikely that $\pi_nhoim(f)=\pi_n(hofib(Y\rightarrow C_f)$ is going to be isomorphic to $f_*\pi_nX$. One case in which it does happen is if $f$ is a weak equivalence. 
In fact the canonical null-homotopy of the composite $X\xrightarrow{f}Y\xrightarrow{k} C_f$ gives you a lift $\overline f:X\rightarrow F_k$
and the Blakers-Massey Excision Theorem tells you that if the map $f$ is $n$-connected and the space $X$ is $m$-connected, then $\overline f$ is $(n+m)$-connected. Thus you see that it is in fact very rare that $\pi_nhomin(f)$ is equal to $f_*\pi_nX$.
Max has supplied one example in the comments of where things fail (since $hofib(S^2\rightarrow S^2\vee S^2)\simeq \Omega S^2\times\Omega(\Omega S^2\ast\Omega S^2)$ you have $\pi_2$ of this fibre is $\mathbb{Z}\oplus\mathbb{Z}$). Here is an example of my own:
Take a space $X$ and let $f$ be the unique map $f:X\rightarrow\ast$. Then we have
$$hoim(f)\simeq \Omega\Sigma X$$
and clearly $\pi_n\Omega\Sigma X\cong\pi_{n+1}\Sigma X$ is not necessarily trivial whilst $f_*\pi_nX\subseteq\{0\}$.
