I've just stumbled on something that seems either too good to be true,
or else too good for me not to have heard of it before.  

It has to do with the basepoint forgetting map
$$
u: [A, M] \to \langle A, M \rangle, 
$$
where $A$ is a pointed space, $M$ is a topological monoid,
and $\langle A, M\rangle$
indicates unpointed homotopy classes.


For general $X$, I like to understand the map $u$
by writing $A\_\circ$ for the unpointed version of $A$,
and setting up the long cofiber sequence
$$
S^0 \to (A\_\circ)\_+ \to A \to S^1 \to \Sigma((A\_\circ)\_+) \to \cdots
$$
where $(A\_\circ)\_+$ indicates the basepoint free $A$ with a disjoint basepoint
added, and $S^0\to (A\_\circ)\_+$ sends the nontrivial point to the basepoint
we've forgotten about. Now we map this into $X$ to get the exact sequence
$$
[S^0, X]\gets[(A\_\circ)\_+, X]\gets [A, X] 
\gets \pi\_1(X) \gets [\Sigma((A\_\circ)\_+), X]\gets\cdots .
$$
The first few terms are just pointed sets, and the map $[(A\_\circ)\_+, X]\gets [A, X]$
can be identified with $u$.


It is easy to see that $\pi\_1(X) \gets [\Sigma((A\_\circ)\_+), X]$
is surjective (it follows from naturality, using the composition $\*\to A\to \*$),
so that the kernel of $u$ is trivial.


However, it is not true that pointed homotopy and unpointed homotopy agree, and this
is captured in the action of $\pi\_1(X)$ on $[A,X]$ coming from the long cofiber sequence.  

We have $u(f) = u(g)$ if and only if $f$ and $g$ are in the same orbit of the action.


The great thing is that when we take $X$ to be a topological monoid $M$, even the
initial terms in the sequence are groups and homomorphisms, so the orbits are
cosets of the image of $[A, M] 
\gets \pi\_1(M)$, which is trivial. Consequently, pointed and unpointed homotopy classes into
$M$ agree.


QUESTION: This fact is not in my mental toolbox! Am I missing something? Alternatively, is there a nice reference to quote for this?



Pointed versus unpointed maps into a topological monoid
As is implicitly pointed out in the comments, you really want to assume that $X$ ($=M$) is path connected. And then your analysis is fine. Note that $M$ will then wish to be equivalent to $\Omega BM$, so those pointed homotopy classes then look like $\langle \Sigma A, BM \rangle$ and fundamental group issues will go away, as $BM$ will be simply connected.


