Let $X = \Bbb{RP}^\infty$. I claim that the map $X \to X^3 / M$ is nontrivial on mod-2 cohomology.

Here is some general material.

For $1 \leq i \leq 3$, let $p_i: X^3 \to X^2$ be the projection away from the $i$th factor, and $A_i$ the inverse image of the diagonal of $X^2$ --- eg $X_1$ is the set of points of the form $(y,x,x)$. All the spaces $A_i$ are homeomorphic to $X^2$, and their common intersections are the diagonal copy of $X$.

Let $P = A_1 \cup A_2 \cup A_3 \subset X^3$. Then there is a continuous map $g: P \to X$ such that $g(x,x,y) = g(x,y,x) = g(y,x,x) = x$, and $X^3 / M$ is the pushout of a diagram $$ X \xleftarrow{g} P \to X^3. $$ The union $P = A_1 \cup A_2 \cup A_3$ is well-behaved enough to apply the Mayer--Vietoris sequence. If we do so, we can calculate that $H^*(P)$ is isomorphic to the set of $\{(f_1, f_2, f_3)\}$ such that $f_i$ is in $H^*(A_i)$ and such that all three $f_i$ restrict to the same element in $H^*(X)$, the cohomology of the diagonal.

We can also apply the Mayer--Vietoris long exact sequence on cohomology to get an exact sequence $$ \dots \to H^*(X^3/M) \to H^*(X) \oplus H^*(X^3) \to H^*(P) \to \dots $$ Therefore, the map $H^*(X^3 / M) \to H^*(X)$ will be nontrivial if there exist a nonzero $f \in H^*(X)$ and a $g \in H^*(Y)$ such that $f$ and $g$ have the same image in $H^*(P)$. Equivalently, we need a nonzero element $f \in H^*(X)$ and a $g \in H^*(X^3)$ such that all the restrictions of $g$ to $H^*(A_i)$ are equal to restrictions of $f$ along the projection $g: A_i \to X$.

In the specific case $X = \Bbb{RP}^\infty$, $H^*(X)$ is the polynomial ring $\Bbb F_2[x]$. Unravelling the above identifications, we are looking for a nonzero polynomial $f(x) \in \Bbb{RP}^\infty$ and an element $g(x_1,x_2,x_3) \in \Bbb F_2[x_1,x_2,x_3] \cong H^*(X^3)$ such that $(g(z,z,x_3), g(z,x_2,z), g(x_1,z,z)) = (f(z), f(z), f(z))$.

In this case we can take $f(x) = x^2$ and $g(x_1,x_2,x_3) = x_1 x_2 + x_2 x_3 + x_3 x_1$, using the fact that we are working mod $2$.

(There is no such pair of single-variable polynomials in characteristic zero.)

Let $X = \Bbb{RP}^n$ for $n \geq 2$. I claim that the map $X \to X^3 / M$ is nontrivial on mod-2 cohomology.

Here is some general material.

For $1 \leq i \leq 3$, let $p_i: X^3 \to X^2$ be the projection away from the $i$th factor, and $A_i$ the inverse image of the diagonal of $X^2$ --- eg $X_1$ is the set of points of the form $(y,x,x)$. All the spaces $A_i$ are homeomorphic to $X^2$, and their common intersections are the diagonal copy of $X$.

Let $P = A_1 \cup A_2 \cup A_3 \subset X^3$. Then there is a continuous map $g: P \to X$ such that $g(x,x,y) = g(x,y,x) = g(y,x,x) = x$, and $X^3 / M$ is the pushout of a diagram $$ X \xleftarrow{g} P \to X^3. $$ The union $P = A_1 \cup A_2 \cup A_3$ is well-behaved enough to apply the Mayer--Vietoris sequence. If we do so, we can calculate that $H^*(P)$ is isomorphic to the set of $\{(f_1, f_2, f_3)\}$ such that $f_i$ is in $H^*(A_i)$ and such that all three $f_i$ restrict to the same element in $H^*(X)$, the cohomology of the diagonal.

We can also apply the Mayer--Vietoris long exact sequence on cohomology to get an exact sequence $$ \dots \to H^*(X^3/M) \to H^*(X) \oplus H^*(X^3) \to H^*(P) \to \dots $$ Therefore, the map $H^*(X^3 / M) \to H^*(X)$ will be nontrivial if there exist a nonzero $f \in H^*(X)$ and a $g \in H^*(Y)$ such that $f$ and $g$ have the same image in $H^*(P)$. Equivalently, we need a nonzero element $f \in H^*(X)$ and a $g \in H^*(X^3)$ such that all the restrictions of $g$ to $H^*(A_i)$ are equal to restrictions of $f$ along the projection $g: A_i \to X$.

In the specific case $X = \Bbb{RP}^{n}$, $H^*(X)$ is the polynomial ring $\Bbb F_2[x]/(x^{n+1})$. Unravelling the above identifications, we are looking for a nonzero polynomial $f(x) \in \Bbb F_2[x]/(x^{n+1})$ and an element $g(x_1,x_2,x_3) \in \Bbb F_2[x_1,x_2,x_3]/(x_i^{n+1}) \cong H^*(X^3)$ such that $(g(z,z,x_3), g(z,x_2,z), g(x_1,z,z)) = (f(z), f(z), f(z))$.

In this case, we can take $f(x) = x^2$ and $g(x_1,x_2,x_3) = x_1 x_2 + x_2 x_3 + x_3 x_1$, using the fact that we are working mod $2$. This is nonzero as long as we are at least at $\Bbb{RP}^2$.

(There is no such pair of single-variable polynomials in characteristic zero.)

Let $X = \Bbb{RP}^\infty$. I claim that the map $X \to X^3 / M$ is nontrivial on mod-2 cohomology.

Here is some general material.

For $1 \leq i \leq 3$, let $\pi_i: X^3 \to X$ be the projection, and $A_i$ the inverse image of the diagonal of $X^2$ --- eg $X_1$ is the set of points of the form $(y,x,x)$. All the spaces $A_i$ are homeomorphic to $X^2$, and their common intersections are the diagonal copy of $X$.

Let $P = A_1 \cup A_2 \cup A_3 \subset X^3$. Then there is a continuous map $g: P \to X$ such that $g(x,x,y) = g(x,y,x) = g(y,x,x) = x$, and $X^3 / M$ is the pushout of a diagram $$ X \xleftarrow{g} P \to X^3. $$ The union $P = A_1 \cup A_2 \cup A_3$ is well-behaved enough to apply the Mayer--Vietoris sequence. If we do so, we can calculate that $H^*(P)$ is isomorphic to the set of $\{(f_1, f_2, f_3)\}$ such that $f_i$ is in $H^*(A_i)$ and such that all three $f_i$ restrict to the same element in $H^*(X)$, the cohomology of the diagonal.

We can also apply the Mayer--Vietoris long exact sequence on cohomology to get an exact sequence $$ \dots \to H^*(X^3/M) \to H^*(X) \oplus H^*(X^3) \to H^*(P) \to \dots $$ Therefore, the map $H^*(X^3 / M) \to H^*(X)$ will be nontrivial if there exist a nonzero $f \in H^*(X)$ and a $g \in H^*(Y)$ such that $f$ and $g$ have the same image in $H^*(P)$. Equivalently, we need a nonzero element $f \in H^*(X)$ and a $g \in H^*(X^3)$ such that all the restrictions of $g$ to $H^*(A_i)$ are equal to restrictions of $f$ along the projection $g: A_i \to X$.

In the specific case $X = \Bbb{RP}^\infty$, $H^*(X)$ is the polynomial ring $\Bbb F_2[x]$. Unravelling the above identifications, we are looking for a nonzero polynomial $f(x) \in \Bbb{RP}^\infty$ and an element $g(x_1,x_2,x_3) \in \Bbb F_2[x_1,x_2,x_3] \cong H^*(X^3)$ such that $(g(z,z,x_3), g(z,x_2,z), g(x_1,z,z)) = (f(z), f(z), f(z))$.

In this case we can take $f(x) = x^2$ and $g(x_1,x_2,x_3) = x_1 x_2 + x_2 x_3 + x_3 x_1$, using the fact that we are working mod $2$.

(There is no such pair of single-variable polynomials in characteristic zero.)

Let $X = \Bbb{RP}^\infty$. I claim that the map $X \to X^3 / M$ is nontrivial on mod-2 cohomology.

Here is some general material.

For $1 \leq i \leq 3$, let $p_i: X^3 \to X^2$ be the projection away from the $i$th factor, and $A_i$ the inverse image of the diagonal of $X^2$ --- eg $X_1$ is the set of points of the form $(y,x,x)$. All the spaces $A_i$ are homeomorphic to $X^2$, and their common intersections are the diagonal copy of $X$.

Let $P = A_1 \cup A_2 \cup A_3 \subset X^3$. Then there is a continuous map $g: P \to X$ such that $g(x,x,y) = g(x,y,x) = g(y,x,x) = x$, and $X^3 / M$ is the pushout of a diagram $$ X \xleftarrow{g} P \to X^3. $$ The union $P = A_1 \cup A_2 \cup A_3$ is well-behaved enough to apply the Mayer--Vietoris sequence. If we do so, we can calculate that $H^*(P)$ is isomorphic to the set of $\{(f_1, f_2, f_3)\}$ such that $f_i$ is in $H^*(A_i)$ and such that all three $f_i$ restrict to the same element in $H^*(X)$, the cohomology of the diagonal.

We can also apply the Mayer--Vietoris long exact sequence on cohomology to get an exact sequence $$ \dots \to H^*(X^3/M) \to H^*(X) \oplus H^*(X^3) \to H^*(P) \to \dots $$ Therefore, the map $H^*(X^3 / M) \to H^*(X)$ will be nontrivial if there exist a nonzero $f \in H^*(X)$ and a $g \in H^*(Y)$ such that $f$ and $g$ have the same image in $H^*(P)$. Equivalently, we need a nonzero element $f \in H^*(X)$ and a $g \in H^*(X^3)$ such that all the restrictions of $g$ to $H^*(A_i)$ are equal to restrictions of $f$ along the projection $g: A_i \to X$.

In the specific case $X = \Bbb{RP}^\infty$, $H^*(X)$ is the polynomial ring $\Bbb F_2[x]$. Unravelling the above identifications, we are looking for a nonzero polynomial $f(x) \in \Bbb{RP}^\infty$ and an element $g(x_1,x_2,x_3) \in \Bbb F_2[x_1,x_2,x_3] \cong H^*(X^3)$ such that $(g(z,z,x_3), g(z,x_2,z), g(x_1,z,z)) = (f(z), f(z), f(z))$.

In this case we can take $f(x) = x^2$ and $g(x_1,x_2,x_3) = x_1 x_2 + x_2 x_3 + x_3 x_1$, using the fact that we are working mod $2$.

(There is no such pair of single-variable polynomials in characteristic zero.)