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The inclusion $\mathbb RP^2\to \mathbb CP^2$ has no kernel in the pointed homotopy category.

Proof:

Suppose $X\to \mathbb RP^2$ is such a kernel. Then maps $S^1\to X$ (in that category) correspond bijectively to maps $S^1\to\mathbb RP^2$ such that the composed map $S^1\to \mathbb CP^2$ is trivial. This means that $\pi_1(X)$ has order $2$.

It follows that there is a map $\mathbb RP^2\to X$ inducing an isomorphism of $\pi_1$. The composed map $\mathbb RP^2\to X\to \mathbb RP^2$ also induces an isomorphism of $\pi_1$, therefore also of $H^1$ with $\mathbb Z/2$ coefficients, therefore (using naturality of cup products) an iso of $H^2$ with $\mathbb Z/2$ coefficients.

But the map $X\to \mathbb RP^2$ must be zero on $H^2$ because the inclusion $\mathbb RP^2\to\mathbb CP^2$ is an isomorphism on $H^2$ and the composed map $X\to \mathbb RP^2 \to \mathbb CP^2$ is zero.

EDIT: Also, the degree two map $S^1\to S^1$ has no cokernel. Briefly, any such cokernel would have to be a retract (up to homotopy) of $\mathbb RP^2$, but that would make it either a point or $\mathbb RP^2$ and neither of these does the job.

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The inclusion $\mathbb RP^2\to \mathbb CP^2$ has no kernel in the pointed homotopy category.

Proof:

Suppose $X\to \mathbb RP^2$ is such a kernel. Then maps $S^1\to X$ (in that category) correspond bijectively to maps $S^1\to\mathbb RP^2$ such that the composed map $S^1\to \mathbb CP^2$ is trivial. This means that $\pi_1(X)$ has order $2$.

It follows that there is a map $\mathbb RP^2\to X$ inducing an isomorphism of $\pi_1$. The composed map $\mathbb RP^2\to X\to \mathbb RP^2$ also induces an isomorphism of $\pi_1$, therefore also of $H^1$ with $\mathbb Z/2$ coefficients, therefore (using naturality of cup products) an iso of $H^2$ with $\mathbb Z/2$ coefficients.

But the map $X\to \mathbb RP^2$ must be zero on $H^2$ because the inclusion $\mathbb RP^2\to\mathbb CP^2$ is an isomorphism on $H^2$ and the composed map $X\to \mathbb RP^2 \to \mathbb CP^2$ is zero.